7 (sec2√x) ((1/2) X – ½). With some experience, you won’t introduce a new variable like $u = \cdots$ as we did above. We have the outer function $f(u) = u^7$ and the inner function $u = g(x) = x^2 +1.$ Then $f'(u) = 7u^6,$ and $g'(x) = 2x.$ Then \begin{align*} f'(x) &= 7u^6 \cdot 2x \\[8px] Identify the mistake(s) in the equation. Differentiate the outer function, ignoring the constant. chain rule example problems MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on CHAIN RULE with easy and logical explanations. This video gives the definitions of the hyperbolic functions, a rough graph of three of the hyperbolic functions: y = sinh x, y = cosh x, y = tanh x Let f(x)=6x+3 and g(x)=−2x+5. d/dx sqrt(x) = d/dx x(1/2) = (1/2) x(-½). By continuing, you agree to their use. In this presentation, both the chain rule and implicit differentiation will &= 8\left(3x^2 – 4x + 5\right)^7 \cdot (6x-4) \quad \cmark \end{align*}. : ), Thank you. It is often useful to create a visual representation of Equation for the chain rule. 1. (2x – 4) / 2√(x2 – 4x + 2). (You don’t need us to show you how to do algebra! Chain Rule - Examples. The number e (Euler’s number), equivalent to about 2.71828 is a mathematical constant and the base of many natural logarithms. Get an idea on partial derivatives-definition, rules and solved examples. Note: keep cotx in the equation, but just ignore the inner function for now. Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. Instead, you’ll think something like: “The function is a bunch of stuff to the 7th power. The second is more formal. D(√x) = (1/2) X-½. • Solution 1. Consider a composite function whose outer function is $f(x)$ and whose inner function is $g(x).$ The composite function is thus $f(g(x)).$ Its derivative is given by: \[\bbox[yellow,8px]{ \begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\], Alternatively, if we write $y = f(u)$ and $u = g(x),$ then \[\bbox[yellow,8px]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\]. &= -\sin(\tan(3x)) \cdot \sec^2 (3x) \cdot 3 \quad \cmark \end{align*}. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. Covered for all Bank Exams, Competitive Exams, Interviews and Entrance tests. : ). Then. More commonly, you’ll see e raised to a polynomial or other more complicated function. Jump down to problems and their solutions. We have the outer function $f(u) = \sqrt{u}$ and the inner function $u = g(x) = x^2 + 1.$ Then $\left(\sqrt{u} \right)’ = \dfrac{1}{2}\dfrac{1}{ \sqrt{u}},$ and $\left(x^2 + 1 \right)’ = 2x.$ Hence \begin{align*} f'(x) &= \dfrac{1}{2}\dfrac{1}{ \sqrt{u}} \cdot 2x \\[8px] We have the outer function $f(z) = \cos z,$ and the middle function $z = g(u) = \tan(u),$ and the inner function $u = h(x) = 3x.$ Then $f'(z) = -\sin z,$ and $g'(u) = \sec^2 u,$ and $h'(x) = 3.$ Hence: \begin{align*} f'(x) &= (-\sin z) \cdot (\sec^2 u) \cdot (3) \\[8px] So the derivative is 3 times that same stuff to the power of 2, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x)\quad \cmark \\[8px] = e5x2 + 7x – 13(10x + 7), Step 4 Rewrite the equation and simplify, if possible. In this case, the outer function is the sine function. &= \sec^2(e^x) \cdot e^x \quad \cmark \end{align*}, Now let’s use the Product Rule: \[ \begin{align*} (f g)’ &= \qquad f’ g\qquad\qquad +\qquad\qquad fg’ \\[8px] We won’t write out all of the tedious substitutions, and instead reason the way you’ll need to become comfortable with: Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies. We’re glad to have helped! Solutions. This 105. is captured by the third of the four branch diagrams on the previous page. Step 2:Differentiate the outer function first. That’s what we’re aiming for. Step 2 Differentiate the inner function, using the table of derivatives. In this example, the negative sign is inside the second set of parentheses. Chain Rule problems or examples with solutions. Need to use the derivative to find the equation of a … &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] h ' ( x ) = 2 ( ln x ) Tip: This technique can also be applied to outer functions that are square roots. Since the functions were linear, this example was trivial. √ (x4 – 37) equals (x4 – 37) 1/2, which when differentiated (outer function only!) Add the constant you dropped back into the equation. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = x^7 – 4x^3 + x.$ Then $f'(u) = e^u,$ and $g'(x) = 7x^6 -12x^2 +1.$ Hence \begin{align*} f'(x) &= e^u \cdot \left(7x^6 -12x^2 +1 \right)\\[8px] dF/dx = dF/dy * dy/dx Step 3: Differentiate the inner function. Label the function inside the square root as y, i.e., y = x2+1. Solution 2 (more formal). &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] The problems below combine the Product rule and the Chain rule, or require using the Chain rule multiple times. \end{align*}. &= 99\left(x^5 + e^x\right)^{98} \cdot \left(5x^4 + e^x\right) \quad \cmark \end{align*}, Solution 2. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)². Click HERE for a real-world example of the chain rule. We’ll solve this two ways. Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. : ), this was really easy to understand good job, Thanks for letting us know. With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. That is _great_ to hear!! Differentiate ``the square'' first, leaving (3 x +1) unchanged. Sample problem: Differentiate y = 7 tan √x using the chain rule. The derivative of ex is ex, so: Solution 1 (quick, the way most people reason). Combine the results from Step 1 (2cot x) (ln 2) and Step 2 ((-csc2)). D(2cot x) = 2cot x (ln 2), Step 2 Differentiate the inner function, which is We have the outer function $f(u) = u^8$ and the inner function $u = g(x) = 3x^2 – 4x + 5.$ Then $f'(u) = 8u^7,$ and $g'(x) = 6x -4.$ Hence \begin{align*} f'(x) &= 8u^7 \cdot (6x – 4) \\[8px] The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. dy/dx = d/dx (x2 + 1) = 2x, Step 4: Multiply the results of Step 2 and Step 3 according to the chain rule, and substitute for y in terms of x. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. The Practically Cheating Calculus Handbook, The Practically Cheating Statistics Handbook, Chain rule examples: Exponential Functions, https://www.calculushowto.com/derivatives/chain-rule-examples/. Want access to all of our Calculus problems and solutions? 7 (sec2√x) ((½) X – ½) = Combine the results from Step 1 (sec2 √x) and Step 2 ((½) X – ½). &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*} We could of course simplify the result algebraically to $14x(x^2+1)^2,$ but we’re leaving the result as written to emphasize the Chain rule term $2x$ at the end. Step 4 &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). In this case, the outer function is x2. Use the chain rule to differentiate composite functions like sin(2x+1) or [cos(x)]³. (a) f(x;y) = 3x+ 4y; @f @x = 3; @f @y = 4. Just ignore it, for now. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] This imaginary computational process works every time to identify correctly what the inner and outer functions are. Step 4: Simplify your work, if possible. The chain rule can be used to differentiate many functions that have a number raised to a power. f’ = ½ (x2 – 4x + 2)½ – 1(2x – 4) Need to review Calculating Derivatives that don’t require the Chain Rule? \text{Then}\phantom{f(x)= }\\ \frac{df}{dx} &= 7(\text{stuff})^6 \cdot \left(\frac{d}{dx}(x^2 + 1)\right) \\[8px] Step 1: Identify the inner and outer functions. In this example, the outer function is ex. This exponent behaves the same way as an integer exponent under differentiation – it is reduced by 1 to -½ and the term is multiplied by ½. Then you would next calculate $10^7,$ and so $(\boxed{\phantom{\cdots}})^7$ is the outer function. Thanks for letting us know! Solution to Example 1. Identify the mistake(s) in the equation. 5x2 + 7x – 19. Step 3. 7 (sec2√x) ((½) 1/X½) = Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(… Tip You can also use this rule to differentiate natural and common base 10 logarithms (D(ln x) = (1/x) and D(log x) = (1/x) log e. Multiplied constants add another layer of complexity to differentiating with the chain rule. x(x2 + 1)(-½) = x/sqrt(x2 + 1). The more times you apply the chain rule to different problems, the easier it becomes to recognize how to apply the rule. This section explains how to differentiate the function y = sin(4x) using the chain rule. Partial derivative is a method for finding derivatives of multiple variables. Combine the results from Step 1 (e5x2 + 7x – 19) and Step 2 (10x + 7). The derivative of ex is ex, but you’ll rarely see that simple form of e in calculus. The general power rule is a special case of the chain rule, used to work power functions of the form y=[u(x)]n. The general power rule states that if y=[u(x)]n], then dy/dx = n[u(x)]n – 1u'(x). The key is to look for an inner function and an outer function. D(4x) = 4, Step 3. If 30 men can build a wall 56 meters long in 5 days, what length of a similar wall can be built … You can find the derivative of this function using the power rule: = (2cot x (ln 2) (-csc2)x). Get complete access: LOTS of problems with complete, clear solutions; tips & tools; bookmark problems for later review; + MORE! -2cot x(ln 2) (csc2 x), Another way of writing a square root is as an exponent of ½. Note: keep 4x in the equation but ignore it, for now. f ' (x) = (df / du) (du / dx) = - 4 sin (u) (5) We now substitute u = 5x - 2 in sin (u) above to obtain. Step 4 Rewrite the equation and simplify, if possible. Step 4 Simplify your work, if possible. Therefore sqrt(x) differentiates as follows: Chain rule Statement Examples Table of Contents JJ II J I Page5of8 Back Print Version Home Page 21.2.6 Example Find the derivative d dx h cos ex4 i. We have Free Practice Chain Rule (Arithmetic Aptitude) Questions, Shortcuts and Useful tips. A simpler form of the rule states if y – un, then y = nun – 1*u’. cot x. Solution: d d x sin( x 2 os( x 2) d d x x 2 =2 x cos( x 2). Step 1 Solution The outside function is the cosine function: d dx h cos ex4 i = sin ex4 d dx h ex4 i = sin ex4 ex4(4x3): The second step required another use of the chain rule (with outside function the exponen-tial function). f ' (x) = - 20 sin (5x - 2) (b) f(x;y) = xy3 + x 2y 2; @f @x = y3 + 2xy2; @f @y = 3xy + 2xy: (c) f(x;y) = x 3y+ ex; @f @x = 3x2y+ ex; @f For example, let’s say you had the functions: The composition g (f (x)), which is also written as (g ∘ f) (x), would be (x2-3)2. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). The chain rule is a rule for differentiating compositions of functions. rule d y d x = d y d u d u d x ecomes Rule) d d x f ( g ( x = f 0 ( g ( x )) g 0 ( x ) \outer" function times of function. The derivative of 2x is 2x ln 2, so: Buy full access now — it’s quick and easy! Once you’ve performed a few of these differentiations, you’ll get to recognize those functions that use this particular rule. For an example, let the composite function be y = √(x4 – 37). = f’ = ½ (x2-4x + 2) – ½(2x – 4), Step 4: (Optional)Rewrite using algebra: In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man." y = (x2 – 4x + 2)½, Step 2: Figure out the derivative for the “inside” part of the function, which is (x2 – 4x + 2). Differentiation Using the Chain Rule SOLUTION 1 : Differentiate. The results are then combined to give the final result as follows: Include the derivative you figured out in Step 1: Worked example: Derivative of ln(√x) using the chain rule. Step 2 Differentiate the inner function, which is The inner function is the one inside the parentheses: x 2 … Hyperbolic Functions And Their Derivatives. equals ½(x4 – 37) (1 – ½) or ½(x4 – 37)(-½). Step 1 Differentiate the outer function, using the table of derivatives. Step 2: Differentiate y(1/2) with respect to y. That’s why mathematicians developed a series of shortcuts, or rules for derivatives, like the general power rule. Chain Rule Example #1 Differentiate $f(x) = (x^2 + 1)^7$. Solution 1 (quick, the way most people reason). In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . Solutions to Examples on Partial Derivatives 1. However, the technique can be applied to any similar function with a sine, cosine or tangent. D(3x + 1)2 = 2(3x + 1)2-1 = 2(3x + 1). : (x + 1)½ is the outer function and x + 1 is the inner function. = cos(4x)(4). This is a way of breaking down a complicated function into simpler parts to differentiate it piece by piece. Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Learn More at BYJU’S. \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. This indicates that the function f(x), the inner function, must be calculated before the value of g(x), the outer function, can be found. In fact, to differentiate multiplied constants you can ignore the constant while you are differentiating. D(5x2 + 7x – 19) = (10x + 7), Step 3. This section shows how to differentiate the function y = 3x + 12 using the chain rule. Now, we just plug in what we have into the chain rule. You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function. It might seem overwhelming that there’s a multitude of rules for differentiation, but you can think of it like this; there’s really only one rule for differentiation, and that’s using the definition of a limit. Step 1 Differentiate the outer function. The outer function in this example is 2x. Knowing where to start is half the battle. Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. Solution: Using the above table and the Chain Rule. Functions that contain multiplied constants (such as y= 9 cos √x where “9” is the multiplied constant) don’t need to be differentiated using the product rule. Let u = cosx so that y = u2 It follows that du dx = −sinx dy du = 2u Then dy dx = dy du × du dx = 2u× −sinx = −2cosxsinx Example Suppose we wish to differentiate y = (2x− 5)10. Remember that a function raised to an exponent of -1 is equivalent to 1 over the function, and that an exponent of ½ is the same as a square root function. Hyperbolic Functions - The Basics. The derivative of x4 – 37 is 4x(4-1) – 0, which is also 4x3. Note that I’m using D here to indicate taking the derivative. AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site. √x. • Solution 1. In this example, 2(3x +1) (3) can be simplified to 6(3x + 1). Step 4: Multiply Step 3 by the outer function’s derivative. In order to use the chain rule you have to identify an outer function and an inner function. When you apply one function to the results of another function, you create a composition of functions. Let’s use the second form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} }\] It’s more traditional to rewrite it as: Worked example: Derivative of √(3x²-x) using the chain rule. Example: Find d d x sin( x 2). The following equation for h ' (x) comes from applying the chain rule incorrectly. For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. Example question: What is the derivative of y = √(x2 – 4x + 2)? We won’t write out “stuff” as we did before to use the Chain Rule, and instead will just write down the answer using the same thinking as above: We can view $\left(x^2 + 1 \right)^7$ as $({\text{stuff}})^7$, where $\text{stuff} = x^2 + 1$. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. Most problems are average. \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= -2(\text{stuff})^{-3} \cdot \dfrac{d}{dx}(\cos x – \sin x) \\[8px] This calculus video tutorial shows you how to find the derivative of any function using the power rule, quotient rule, chain rule, and product rule. Step 2: Differentiate the inner function. For example, to differentiate In this example, cos(4x)(4) can’t really be simplified, but a more traditional way of writing cos(4x)(4) is 4cos(4x). Please read and accept our website Terms and Privacy Policy to post a comment. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Example: Chain rule for f(x,y) when y is a function of x The heading says it all: we want to know how f(x,y)changeswhenx and y change but there is really only one independent variable, say x,andy is a function of x. ), Solution 2 (more formal). Chain Rule Practice Problems: Level 01 Chain Rule Practice Problems : Level 02 If 10 men or 12 women take 40 days to complete a piece of work, how long … Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. &= \dfrac{1}{2}\dfrac{1}{ \sqrt{x^2+1}} \cdot 2x \quad \cmark \end{align*}, Solution 2 (more formal). In this example, the inner function is 3x + 1. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Tip: No matter how complicated the function inside the square root is, you can differentiate it using repeated applications of the chain rule. Step 3: Combine your results from Step 1 2(3x+1) and Step 2 (3). Step 3. Show Solution. The second is more formal. Technically, you can figure out a derivative for any function using that definition. Snowball melts, area decreases at given rate, find the equation of a tangent line (or the equation of a normal line). There are lots more completely solved example problems below! So the derivative is $-2$ times that same stuff to the $-3$ power, times the derivative of that stuff.” \[ \bbox[10px,border:2px dashed blue]{\dfrac{df}{dx} = \left[\dfrac{df}{d\text{(stuff)}}\text{, with the same stuff inside} \right] \times \dfrac{d}{dx}\text{(stuff)}}\] Examples az ax ; az ду ; 2. Watch the video for a couple of chain rule examples, or read on below: The formal definition of the chain rule: Solution 2 (more formal). Just ignore it, for now. The first is the way most experienced people quickly develop the answer, and that we hope you’ll soon be comfortable with. Differentiating using the chain rule usually involves a little intuition. In integration, the counterpart to the chain rule is the substitution rule . Let u = 5x - 2 and f (u) = 4 cos u, hence. Step 1: Rewrite the square root to the power of ½: (10x + 7) e5x2 + 7x – 19. It is useful when finding the … Example problem: Differentiate the square root function sqrt(x2 + 1). We now use the chain rule. That material is here. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] Recall that $\dfrac{d}{du}\left(u^n\right) = nu^{n-1}.$ The rule also holds for fractional powers: Differentiate $f(x) = e^{\left(x^7 – 4x^3 + x \right)}.$. To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. &= 7(x^2+1)^6 \cdot 2x \quad \cmark \end{align*}. We’ll illustrate in the problems below. Compute the integral IS zdrdyd: if D is bounded by the surfaces: D 4. D(e5x2 + 7x – 19) = e5x2 + 7x – 19. Example problem: Differentiate y = 2cot x using the chain rule. The outer function in this example is “tan.” (Note: Leave the inner function in the equation (√x) but ignore that too for the moment) The derivative of tan x is sec2x, so: = 2(3x + 1) (3). No other site explains this nice. The derivative of cot x is -csc2, so: Get notified when there is new free material. 1. Both use the rules for derivatives by applying them in slightly different ways to differentiate the complex equations without much hassle. &= -2(\cos x – \sin x)^{-3} \cdot (-\sin x – \cos x) \quad \cmark \end{align*} We could simplify the answer by factoring out the negative signs from the last term, but we prefer to stop there to keep the focus on the Chain rule. 2x * (½) y(-½) = x(x2 + 1)(-½), Step 5: Simplify your answer by writing it in terms of square roots. Let’s first think about the derivative of each term separately. We’ll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that’s the one you’ll use to compute derivatives quickly as the course progresses. Want to skip the Summary? How can I tell what the inner and outer functions are? The chain rule in calculus is one way to simplify differentiation. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \\[8px] The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). • Solution 2. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] (The outer layer is ``the square'' and the inner layer is (3 x +1). Although it’s tedious to write out each separate function, let’s use an extension of the first form of the Chain rule above, now applied to $f\Bigg(g\Big(h(x)\Big)\Bigg)$: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Bigg(g\Big(h(x)\Big)\Bigg) \right]’ &= f’\Bigg(g\Big(h(x)\Big)\Bigg) \cdot g’\Big(h(x)\Big) \cdot h'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the middle function] } \\[5px]&\qquad \times \text{ [derivative of the middle function, evaluated at the inner function]} \\[5px]&\qquad \quad \times \text{ [derivative of the inner function]}\end{align*}}\] &= 7(x^2 + 1)^6 \cdot (2x) \quad \cmark \end{align*} Note: You’d never actually write “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. Step 1 Differentiate the outer function first. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}}\] We’ll again solve this two ways. : ), What a great site. We’re glad you found them good for practicing. Solutions. We have the outer function $f(u) = u^{-2}$ and the inner function $u = g(x) = \cos x – \sin x.$ Then $f'(u) = -2u^{-3},$ and $g'(x) = -\sin x – \cos x.$ (Recall that $(\cos x)’ = -\sin x,$ and $(\sin x)’ = \cos x.$) Hence \begin{align*} f'(x) &= -2u^{-3} \cdot (-\sin x – \cos x) \\[8px] Differentiate $f(x) = (\cos x – \sin x)^{-2}.$, Differentiate $f(x) = \left(x^5 + e^x\right)^{99}.$. D(sin(4x)) = cos(4x). √ X + 1  Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. The Chain Rule is a big topic, so we have a separate page on problems that require the Chain Rule. &= 3\big[\tan x\big]^2 \cdot \sec^2 x \\[8px] In this example, the inner function is 4x. The inner function is the one inside the parentheses: x4 -37. Huge thumbs up, Thank you, Hemang! \left[\left(x^2 + 1 \right)^7 (3x – 7)^4 \right]’ &= \left[ \left(x^2 + 1 \right)^7\right]’ (3x – 7)^4\, + \,\left(x^2 + 1 \right)^7 \left[(3x – 7)^4 \right]’ \\[8px] What’s needed is a simpler, more intuitive approach! f'(x2 – 4x + 2)= 2x – 4), Step 3: Rewrite the equation to the form of the general power rule (in other words, write the general power rule out, substituting in your function in the right places). Solution. We have the outer function $f(u) = \sin u$ and the inner function $u = g(x) = 2x.$ Then $f'(u) = \cos u,$ and $g'(x) = 2.$ Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] \text{Then}\phantom{f(x)= }\\ \dfrac{df}{dx} &= 3\big[\text{stuff}\big]^2 \cdot \dfrac{d}{dx}(\tan x) \\[8px] —– We could of course simplify this expression algebraically: $$f'(x) = 14x\left(x^2 + 1 \right)^6 (3x – 7)^4 + 12 \left(x^2 + 1 \right)^7 (3x – 7)^3 $$ We instead stopped where we did above to emphasize the way we’ve developed the result, which is what matters most here. Think something like: “The function is some stuff to the $-2$ power. The outer function is √, which is also the same as the rational exponent ½. Hint : Recall that with Chain Rule problems you need to identify the “ inside ” and “ outside ” functions and then apply the chain rule. Chain rule for partial derivatives of functions in several variables. \begin{align*} f(x) &= (\text{stuff})^7; \quad \text{stuff} = x^2 + 1 \\[12px] CHAIN RULE MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Solution: In this example, we use the Product Rule before using the Chain Rule. Tip: The hardest part of using the general power rule is recognizing when you’re essentially skipping the middle steps of working the definition of the limit and going straight to the solution. We’re happy to have helped! We have the outer function $f(u) = u^{99}$ and the inner function $u = g(x) = x^5 + e^x.$ Then $f'(u) = 99u^{98},$ and $g'(x) = 5x^4 + e^x.$ Hence \begin{align*} f'(x) &= 99u^{98} \cdot (5x^4 + e^x) \\[8px] In this example, no simplification is necessary, but it’s more traditional to write the equation like this: \begin{align*} f(x) &= \big[\text{stuff}\big]^3; \quad \text{stuff} = \tan x \\[12px] Multiplying 4x3 by ½(x4 – 37)(-½) results in 2x3(x4 – 37)(-½), which when worked out is 2x3/(x4 – 37)(-½) or 2x3/√(x4 – 37). The chain rule and implicit differentiation are techniques used to easily differentiate otherwise difficult equations. We have the outer function $f(u) = \tan u$ and the inner function $u = g(x) = e^x.$ Then $f'(u) = \sec^2 u,$ and $g'(x) = e^x.$ Hence \begin{align*} f'(x) &= \sec^2 u \cdot e^x \\[8px] Your first 30 minutes with a Chegg tutor is free! That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g. We use cookies to provide you the best possible experience on our website.