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 basic derivative rules power rule product rule quotient rule Some of the practice problems require you to know one or more of the following rules. If you haven't learned all these rules yet, no worries. You can filter out the problems that require techniques you do not know. trig derivatives exponential derivatives derivatives of logarithms

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Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
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 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
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 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

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 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
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17calculus > derivatives > chain rule

 What The Chain Rule Says Substitution Method Direct Method Practice

The Chain Rule is probably the most important derivative rule that you will learn since you will need to use it a lot and it shows up in various forms in other derivatives and integration. It can also be a little confusing at first but if you stick with it, you will be able to understand it well.

There are two main ways to learn the chain rule, substitution and direct. I recommend writing out the substitution form while you are first learning. After you get the hang of it, you can use the direct (shorter) version.
NOTE: Do not be lazy here and try to use the direct version first. Write everything out so that you can see it. If you don't, you will struggle with this concept for the rest of your calculus life and this rule is used everywhere, and I mean EVERYWHERE from now on. You can't get away from it and you can't get around it. So, let's get started.

What The Chain Rule Says

The chain rule says, if you have a function in the form y=f(u) where u is a function of x, then $$\displaystyle{ \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} }$$.
The notation tells you that $$u(x)$$ is a composite function of $$f$$.

Substitution Method

This is the method you should learn first since it breaks the problem into smaller, bite-size pieces that you already know how to take derivatives of. The best way to learn this concept is with examples.

1. $$y=(x+1)^2$$ can be separated into $$u = x+1$$ and $$y=u^2$$. We started on the outside and noticed that there was something with a power of 2. So we took that something and set it equal to u.
2. $$y=\sin(x^2)$$ can be separated into $$u=x^2$$ and $$y=\sin(u)$$. We noticed that, on the outside, we have the sine function. So we set whatever was on the inside to u.
3. $$y=\sqrt{\sin(x)}$$ has an outside function as a square root, while the inside function is sine. So $$u=\sin(x)$$ and $$y = \sqrt{u}$$.
4. $$y=\sqrt{\sin(x^2)}$$ is a combination of the last two examples and is a little more complicated (but not much). We do the same thing as before and notice that the outside function a square root. So we set u to the inside function, i.e. $$u=\sin(x^2)$$ and $$y=\sqrt{u}$$. When we try to take the derivative $$du/dx,$$ we need to do substitution again. So we let $$v=x^2$$ and $$u=\sin(v)$$.

In all of these examples, we just showed you how to set the problem with substitution. You would then complete the problem by taking the derivative of each piece. Try this example before going on to the direct method.

Calculate the derivative of $$y=\cos(x^2+3)$$ using the substitution method.

$$dy/dx = (-2x)\sin(x^2+3)$$

Problem Statement - Calculate the derivative of $$y=\cos(x^2+3)$$ using the substitution method.
Solution - Using the substitution $$u=x^2+3$$, $$y=\cos(x^2+3)$$ can be written $$y=\cos(u)$$.
$$\begin{array}{rcl} \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{dy}{du} \cdot \frac{du}{dx}} \\ & = & \displaystyle{\frac{d}{du}[\cos(u)] \cdot \frac{d}{dx}[x^2+3] }\\ & = & (-\sin(u)) \cdot (2x) \\ & = & (-2x)\sin(x^2+3) \end{array}$$

$$dy/dx = (-2x)\sin(x^2+3)$$

Direct Method

Okay, once you get the basic idea of the chain rule down, the next step is be able to write out the derivative without using substitution. Note: Some instructors jump to this technique right away without emphasizing the substitution concept. If your instructor does that, check your textbook and learn the chain rule with substitution first. You will be glad you did later on down the road.

Now, there are two ways to think about working problems without substitution. The technique that I think works best is to start from the outside and work your way in. Let's see how this is done using an example of an outside to inside technique.

Calculate the derivative of $$g(t)=\sin(t^3)$$ using the direct method.

$$\displaystyle{ g'(t) = (3t^2)\cos(t^3) }$$

Problem Statement - Calculate the derivative of $$g(t)=\sin(t^3)$$ using the direct method.
Solution - The outside function is sine. So we write $$\displaystyle{ \frac{dg}{dt} = \cos(t^3) \cdot \frac{d}{dt}[t^3] }$$. Notice here that we wrote the derivative of the outside and left the inside intact, i.e. the $$t^3$$ inside did not change in the term $$\cos(t^3)$$. Then the second term is the derivative of the inside, written as $$\displaystyle{ \frac{d}{dt}[t^3] }$$. Finishing this example, the answer is

$$\displaystyle{ g'(t) = (3t^2)\cos(t^3) }$$

A second way to work this is to start on the inside and work your way out. Some of the videos further down on this page show this technique but we do not recommend it. The reason is that the more difficult problems (and ones you will probably see on your exam) are nested, i.e. you need to use the chain rule repeatedly. It is much easier to break the problem down and see what you need to do if you start on the outside and work your way in. Starting in the inside is sometimes difficult to find where to start. Let's do an example of this.

Evaluate $$\displaystyle{ \frac{d}{dx}[\sin(\tan(2x))]}$$.

$$\displaystyle{ \frac{d}{dx}[\sin(\tan(2x))] = 2 ~ \cos(\tan(2x)) ~ \sec^2 (2x)}$$

Problem Statement - Evaluate $$\displaystyle{ \frac{d}{dx}[\sin(\tan(2x))]}$$.
Solution - This looks harder than it really is. If you follow the outside-in technique that we just mentioned and you write out some intermediate steps, you should be able to work this kind of problem easily.

 $$\displaystyle{ \frac{d}{dx}[\sin(\tan(2x))]}$$ $$\displaystyle{\cos(\tan(2x)) \cdot \frac{d}{dx}[ \tan(2x) ]}$$ $$\displaystyle{\cos(\tan(2x))\sec^2(2x) \cdot \frac{d}{dx}[2x]}$$ $$\cos(\tan(2x)) ~ \sec^2(2x) ~ 2$$ $$2 \cos(\tan(2x)) ~ \sec^2 (2x)$$

Let's look at each step individually.
In the term $$\displaystyle{ \sin(\tan(2x)) }$$, the outside function is sine, the inside function is $$\tan(2x)$$.
So the term $$\cos(\tan(2x))$$ is the derivative of the outside function (sine) carrying along the inside term without changing it.
Next, we take the derivative of the inside term $$\displaystyle{ \frac{d}{dx}[ \tan(2x) ] }$$.
Now, in this second term there is an outside function, tangent, and an inside function, $$2x$$. The second line takes the derivative of $$\tan(2x)$$ which is $$\sec^2(2x)$$, keeping the inside term intact and then taking the derivative of the inside function, $$d[2x]/dx = 2$$.

$$\displaystyle{ \frac{d}{dx}[\sin(\tan(2x))] = 2 ~ \cos(\tan(2x)) ~ \sec^2 (2x)}$$

Before working practice problems, let's watch a video. This video has a great explanation of the chain rule at the first and then some examples that should help you understand how to use this rule.

### PatrickJMT - Chain Rule for Finding Derivatives [4min-14secs]

video by PatrickJMT

Okay, now you are ready for some practice problems.

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Derivative Chain Rule - Practice Problems Conversion

[A01-973] - [A02-974] - [A03-975] - [A04-976] - [A05-979] - [A06-980] - [A07-992] - [A08-985] - [A09-1907]

[A10-1945] - [A11-983] - [A12-984] - [A13-981] - [A14-991] - [A15-1940] - [A16-2079]

[B01-982] - [B02-986] - [B03-987] - [B04-1310] - [B05-1923] - [B06-1939] - [C01-988] - [C02-990] - [C03-989]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, calculate the derivative of these functions and give your final answers in completely factored form.
By now you probably have learned the derivative of trig functions, exponentials and logarithms. So we do not separate them here. However, if you haven't learned those derivatives yet, it is easy to tell which problems you need to skip.

Basic Problems

$$f(x)=(x+1)^3$$

Problem Statement

Use the chain rule to calculate the derivative of $$f(x)=(x+1)^3$$.

$$3(x+1)^2$$

Problem Statement

Use the chain rule to calculate the derivative of $$f(x)=(x+1)^3$$.

Solution

### 980 solution video

video by Krista King Math

$$3(x+1)^2$$

$$\displaystyle{ (x^3+1)^4 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ (x^3+1)^4 }$$.

$$\displaystyle{ \frac{d}{dx} \left[ (x^3+1)^4 \right] = 12x^2(x^3+1)^3 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ (x^3+1)^4 }$$.

Solution

 $$\displaystyle{ \frac{d}{dx} \left[ (x^3+1)^4 \right] }$$ $$\displaystyle{ 4(x^3+1)^3 \cdot \frac{d}{dx}[x^3+1] }$$ $$4(x^3+1)^3 \cdot (3x^2)$$ $$12x^2(x^3+1)^3$$ Another way to work this is with the substitution method. let $$u=x^3+1$$ $$\displaystyle{ \frac{d}{dx} \left[ (x^3+1)^4 \right] }$$ $$\displaystyle{ 3\frac{d[u^4]}{du} \cdot \frac{d[x^3+1]}{dx} }$$ $$4u^3 \cdot (3x^2)$$ $$12x^2(x^3+1)^3$$

$$\displaystyle{ \frac{d}{dx} \left[ (x^3+1)^4 \right] = 12x^2(x^3+1)^3 }$$

$$\displaystyle{ f(x)=\sqrt{2x+1} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ f(x)=\sqrt{2x+1} }$$.

$$\displaystyle{\frac{1}{\sqrt{2x+1}}}$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ f(x)=\sqrt{2x+1} }$$.

Solution

### 979 solution video

video by Krista King Math

$$\displaystyle{\frac{1}{\sqrt{2x+1}}}$$

$$7 \sin(x^2+1)$$

Problem Statement

Use the chain rule to calculate the derivative of $$7 \sin(x^2+1)$$.

$$(14x) \cos(x^2+1)$$

Problem Statement

Use the chain rule to calculate the derivative of $$7 \sin(x^2+1)$$.

Solution

 $$\displaystyle{ \frac{d}{dx}[ 7\sin(x^2+1) ] }$$ Use the constant multiple rule to move the 7 outside the derivative. $$\displaystyle{ 7\frac{d}{dx}[\sin(x^2+1)] }$$ Apply the chain rule. $$\displaystyle{ 7\cos(x^2+1)\frac{d}{dx}[x^2+1] }$$ Take the derivative of the last term. $$7\cos(x^2+1)(2x)$$ Simplify. $$(14x)\cos(x^2+1)$$

A more explicit way to work this is to use substitution, as follows.

 $$\displaystyle{ \frac{d}{dx}[ 7\sin(x^2+1) ] }$$ Let $$u = x^2+1$$ which is the inside term. $$\displaystyle{ 7\frac{d}{du}[\sin(u)] \cdot \frac{d}{dx}[x^2+1] }$$ $$7[\cos(u)](2x)$$ Make sure and replace the u with $$x^2+1$$, so that you don't have any u's in your final answer. And finally, simplify. $$(14x)\cos(x^2+1)$$

If we call the original function, say $$f(x)=7\sin(x^2+1)$$, the first line is $$\displaystyle{ \frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} }$$ using the Chain Rule. This is probably a better way to work the problem than the first solution, especially while you are learning. Once you have the chain rule down, you can easily work it as shown in the first solution.

$$(14x) \cos(x^2+1)$$

$$\displaystyle{ 3e^{ x^2+7 } }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ 3e^{ x^2+7 } }$$.

$$\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] = 6xe^{x^2+7} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ 3e^{ x^2+7 } }$$.

Solution

 $$\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] }$$ $$\displaystyle{ 3e^{x^2+7} \cdot \frac{d}{dx}[x^2+7] }$$ $$3e^{x^2+7} \cdot (2x)$$ $$6xe^{x^2+7}$$

Another way to work this is with the substitution method.

 let $$u=x^2+7$$ $$\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] }$$ $$\displaystyle{ 3\frac{d[e^u]}{du} \cdot \frac{d[x^2+7]}{dx} }$$ $$3e^u \cdot (2x)$$ $$6xe^{x^2+7}$$

$$\displaystyle{ \frac{d}{dx} \left[ 3e^{ x^2+7 } \right] = 6xe^{x^2+7} }$$

$$\displaystyle{ \ln(3x^2+9x-5) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ \ln(3x^2+9x-5) }$$.

$$\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ \ln(3x^2+9x-5) }$$.

Solution

 $$\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }$$ Apply the chain rule. $$\displaystyle{ \frac{1}{3x^2+9x-5}\cdot \frac{d}{dx}[3x^2+9x-5] }$$ $$\displaystyle{ \frac{1}{3x^2+9x-5}\cdot (6x+9) }$$ $$\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }$$

Using the substitution $$u = 3x^2+9x-5$$ this could have solved more explicitly.

 $$\displaystyle{ \frac{d}{dx}[ \ln(3x^2+9x-5) ] }$$ $$\displaystyle{ \frac{d}{du}[\ln(u)] \cdot \frac{d}{dx}[3x^2+9x-5] }$$ $$\displaystyle{ \frac{1}{u} \cdot (6x+9) }$$ $$\displaystyle{ \frac{1}{3x^2+9x-5} \cdot [ 3(2x+3) ] }$$

$$\displaystyle{ \frac{3(2x+3)}{3x^2+9x-5} }$$

Calculate the first three derivatives of $$\displaystyle{ f(x)=\ln(2+3x) }$$.

Problem Statement

Calculate the first three derivatives of $$\displaystyle{ f(x)=\ln(2+3x) }$$.

Solution

First Derivative

 $$\displaystyle{ f'(x) = \frac{d}{dx}[ \ln(2+3x) ] }$$ $$\displaystyle{\frac{1}{2+3x} \frac{d}{dx}[2+3x]}$$ $$\displaystyle{\frac{1}{2+3x} (3) = \frac{3}{2+3x}}$$

Second Derivative
We can use the either quotient rule or product rule here. The product rule is the easiest here because of the constant in the numerator but we need to rewrite the first derivative as $$\displaystyle{ \frac{3}{2+3x} = 3(2+3x)^{-1} }$$.

 $$\displaystyle{ f''(x) = \frac{d}{dx} \left[ 3(2+3x)^{-1} \right] }$$ $$\displaystyle{3(-1)(2+3x)^{-2} \frac{d}{dx}[2+3x] }$$ $$\displaystyle{-3(2+3x)^{-2} (3) = \frac{-9}{(2+3x)^{2}} }$$

Third Derivative
Again, we can use either the quotient rule or product rule and we choose the product rule.

 $$f^{(3)}(x) = \displaystyle{\frac{d}{dx}\left[ -9(2+3x)^{-2} \right]}$$ $$\displaystyle{-9(-2)(2+3x)^{-3}\frac{d}{dx}[2+3x]}$$ $$18(2+3x)^{-3} (3)$$ $$\displaystyle{\frac{54}{(2+3x)^3}}$$

$$\displaystyle{f'(x) =\frac{3}{2+3x}}$$

$$\displaystyle{f''(x)=\frac{-9}{(2+3x)^{2}}}$$

$$\displaystyle{f^{(3)}(x)=\frac{54}{(2+3x)^3}}$$

$$f(x) = (x^4+3x^2-2)^5$$

Problem Statement

Use the chain rule to calculate the derivative of $$f(x) = (x^4+3x^2-2)^5$$.

Solution

### 985 solution video

video by Krista King Math

$$\tan(x^5 + 2x^3 - 12x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$\tan(x^5 + 2x^3 - 12x)$$.

$$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] = (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\tan(x^5 + 2x^3 - 12x)$$.

Solution

Using substitution, the outside function is tangent and the inside function is everything inside the tangent parentheses. So, we let $$u = x^5 + 2x^3 - 12x$$.

 $$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] }$$ $$\displaystyle{ \frac{d}{dx}[\tan(u)] }$$ $$\displaystyle{ \frac{d}{du}[\tan(u)] \frac{d}{dx}[u] }$$ $$\displaystyle{ \frac{d}{du}[\tan(u)] \frac{d}{dx}[x^5 + 2x^3 - 12x] }$$ $$\sec^2(u) (5x^4 + 6x^2 - 12)$$ $$(5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x)$$

$$\displaystyle{ \frac{d}{dx}[\tan(x^5 + 2x^3 - 12x)] = (5x^4 + 6x^2 - 12) \sec^2(x^5 + 2x^3 - 12x) }$$

$$y=4^{x^3-\sin x}$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=4^{x^3-\sin x}$$.

Solution

### 1945 solution video

video by PatrickJMT

$$y=\sin^3(x) \tan(4x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=\sin^3(x) \tan(4x)$$.

Solution

### 981 solution video

video by Krista King Math

$$g(x)= (1+4x)^5(3+x-x^2)^8$$

Problem Statement

Use the chain rule to calculate the derivative of $$g(x)= (1+4x)^5(3+x-x^2)^8$$.

Solution

### 983 solution video

video by Krista King Math

$$f(x) = (2x-3)^4 (x^2+x+1)^5$$

Problem Statement

Use the chain rule to calculate the derivative of $$f(x) = (2x-3)^4 (x^2+x+1)^5$$.

Solution

### 984 solution video

video by Krista King Math

$$y = (2x^5-3) \sin(7x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (2x^5-3) \sin(7x)$$.

$$y' = 7(2x^5-3)\cos(7x) + 10x^4\sin(7x)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (2x^5-3) \sin(7x)$$.

Solution

 $$\displaystyle{ \frac{dy}{dx} = (2x^5-3) \frac{d}{dx}[\sin(7x)] + \sin(7x)\frac{d}{dx} [2x^5-3] }$$ $$y' = (2x^5-3)\cos(7x) (7) + \sin(7x) (10x^4)$$ $$y' = 7(2x^5-3)\cos(7x) + (10x^4)\sin(7x)$$

$$y' = 7(2x^5-3)\cos(7x) + 10x^4\sin(7x)$$

$$y=x^2\cos(1/x^3)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=x^2\cos(1/x^3)$$.

$$y'=2x\cos(1/x^3)+3\sin(1/x^3)/x^2$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=x^2\cos(1/x^3)$$.

Solution

### 1940 solution video

video by PatrickJMT

$$y'=2x\cos(1/x^3)+3\sin(1/x^3)/x^2$$

$$(x^2-1)e^{-x}$$

Problem Statement

Use the chain rule to calculate the derivative of $$(x^2-1)e^{-x}$$.

$$e^{-x}(1+2x-x^2)$$

Problem Statement

Use the chain rule to calculate the derivative of $$(x^2-1)e^{-x}$$.

Solution

### 2079 solution video

video by Krista King Math

$$e^{-x}(1+2x-x^2)$$

Calculate the third derivative of $$f(x)=\tan(3x)$$.

Problem Statement

Calculate the third derivative of $$f(x)=\tan(3x)$$.

$$54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]$$

Problem Statement

Calculate the third derivative of $$f(x)=\tan(3x)$$.

Solution

$$f'(x)=3\sec^2(3x)$$
$$f''(x)=18\sec^2(3x)\tan(3x)$$

### 2085 solution video

video by PatrickJMT

$$54\sec^2(3x)\left[ 2\tan^2(3x)+\sec^2(3x)\right]$$

Intermediate Problems

$$y=\cot^4(x/2)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=\cot^4(x/2)$$.

$$y'=-2\cot^3(x/2)\csc^2(x/2)$$

Problem Statement

Use the chain rule to calculate the derivative of $$y=\cot^4(x/2)$$.

Solution

### 1910 solution video

video by Krista King Math

$$y'=-2\cot^3(x/2)\csc^2(x/2)$$

$$\displaystyle{ (2x)(x^3+7)^{10} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ (2x)(x^3+7)^{10} }$$.

$$\displaystyle{ 2 (x^3+7)^9 (31x^3+7) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ (2x)(x^3+7)^{10} }$$.

Solution

Starting on the outside and working in, notice that the first things we come across are two factors multiplied together. The two factors are $$(2x)$$ and $$(x^3+7)^{10}$$. So our first step is to use the product rule. Just to make our notation a bit easier, let's call the original function $$g(x)$$, i.e. let $$g(x)=(2x)(x^3+7)^{10}$$.

product rule

$$\displaystyle{ \frac{d}{dx}[g(x)] = \frac{d}{dx} \left[ (2x)(x^3+7)^{10} \right] }$$

$$\displaystyle{ 2x \frac{d}{dx}\left[ (x^3+7)^{10} \right] + (x^3+7)^{10} \frac{d}{dx}[2x] }$$

Okay, so the derivative $$\displaystyle{\frac{d}{dx}[2x] = 2}$$ is pretty easy. This makes the second term $$2(x^3+7)^{10}$$. Now let's look more closely at the derivative in the first term. This derivative requires the chain rule.

chain rule

$$\displaystyle{ \frac{d}{dx}\left[ (x^3+7)^{10} \right] }$$

$$\displaystyle{ 10(x^3+7)^9 \frac{d}{dx}[x^3+7] }$$

$$10(x^3+7)^9 (3x^2)$$

Notice in this step we took the derivative of the outside to get the $$10(x^3+7)^9$$ term and multiplied that by the derivative of the inside term to get $$(3x^2)$$. This is the application of the chain rule.
Let's put all the terms back together and simplify a bit.

simplifying

$$\displaystyle{ \frac{d}{dx}[g(x)] = (2x)[10(x^3+7)^9 (3x^2)]+2(x^3+7)^{10} }$$

$$(60x^3)(x^3+7)^9+2(x^3+7)^{10}$$

At this point, you need to know what your instructor expects as far as simplifying. We usually ask our students to factor, so that is what we will do here. First, let's separate out the terms we have in common in the two factors. We can tell that we have a $$2$$ in both terms and we also have $$(x^3+7)$$. In the second term we have $$10$$ but in the first we have only $$9$$, so we can factor out only $$9$$, i.e. $$(x^3+7)^9$$. So our common factor in both terms is $$2(x^3+7)^9$$. In this last step, we factor and simplify the remaining terms to get our final answer.

factoring and simplifying

$$\displaystyle{ \frac{d}{dx}[g(x)] = (60x^3)(x^3+7)^9+2(x^3+7)^{10} }$$

$$[2(x^3+7)^9](30x^3)+[2(x^3+7)^9](x^3+7)$$

$$2(x^3+7)^9[30x^3+(x^3+7)]$$

$$2(x^3+7)^9(31x^3+7)$$

A Note About Simplifying - - When using the derivative rules, it pays to stop at each step and see what can be factored before going on. Knowing when to do this will become easier with experience. But it is always a good rule of thumb to see what can be factored BEFORE multiplying out a bunch of terms.

$$\displaystyle{ 2 (x^3+7)^9 (31x^3+7) }$$

$$\displaystyle{ \left[ (2x)(x^3+7) \right]^5 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ \left[ (2x)(x^3+7) \right]^5 }$$.

$$\displaystyle{ 160x^4 (x^3+7)^4 (4x^3+7) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ \left[ (2x)(x^3+7) \right]^5 }$$.

Solution

 $$\displaystyle{ \frac{d}{dx}\left[ (2x)(x^3+7) \right]^5 }$$ Apply the chain rule. $$\displaystyle{ 5 \left[ (2x)(x^3+7) \right]^4 \frac{d}{dx} \left[ (2x)(x^3+7) \right] }$$ Use the product rule. $$\displaystyle{ 5 \left[ (2x)(x^3+7) \right]^4 \left[ 2x(3x^2) + (x^3+7)2 \right] }$$ $$10 \left[ (2x)(x^3+7) \right]^4 (3x^3 + x^3 +7)$$ $$10 (2x)^4 (x^3+7)^4 (4x^3+7)$$ $$160x^4 (x^3+7)^4 (4x^3+7)$$

$$\displaystyle{ 160x^4 (x^3+7)^4 (4x^3+7) }$$

$$y = (1+\cos^2(7x))^3$$

Problem Statement

Use the chain rule to calculate the derivative of $$y = (1+\cos^2(7x))^3$$.

Solution

### 982 solution video

video by Krista King Math

$$\displaystyle{ y = \cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right) }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y = \cos \left( \frac{1-e^{2x}}{1+e^{2x}} \right) }$$.

Solution

### 986 solution video

video by Krista King Math

$$\displaystyle{ y=\left( \frac{x^2+1}{x^2-1} \right)^3 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y=\left( \frac{x^2+1}{x^2-1} \right)^3 }$$.

Solution

### 987 solution video

video by Krista King Math

$$\displaystyle{ g(t) = \frac{(t+4)^{1/2}}{(t-4)^{1/2}} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ g(t) = \frac{(t+4)^{1/2}}{(t-4)^{1/2}} }$$.

$$\displaystyle{g'(t) =\frac{-4}{(t-4)^{3/2}(t+4)^{1/2}}}$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ g(t) = \frac{(t+4)^{1/2}}{(t-4)^{1/2}} }$$.

Solution

There is a mistake in this video. Watch the video and try to find it before reading on.

I hope you found the mistake yourself. When he simplifies $$[ (t-4)^1 - (t+4)^1]$$ the answer should be $$[ (t-4)^1 - (t+4)^1] = [ t-4 -t-4 ] = -8$$. In the video he has $$-16$$. So the final answer is $$\displaystyle{ \frac{-4}{(t-4)^{3/2}(t+4)^{1/2}}}$$

### 1310 solution video

video by PatrickJMT

$$\displaystyle{g'(t) =\frac{-4}{(t-4)^{3/2}(t+4)^{1/2}}}$$

Use the quotient rule to find $$f'(x)$$ of $$\displaystyle{ f(x)=\frac{6}{\ln(8x^2)} }$$.

Problem Statement

Use the quotient rule to find $$f'(x)$$ of $$\displaystyle{ f(x)=\frac{6}{\ln(8x^2)} }$$.

$$\displaystyle{f'(x)=\frac{-12}{x[\ln(8x^2)]^2}}$$

Problem Statement

Use the quotient rule to find $$f'(x)$$ of $$\displaystyle{ f(x)=\frac{6}{\ln(8x^2)} }$$.

Solution

### 1923 solution video

video by MathTV

$$\displaystyle{f'(x)=\frac{-12}{x[\ln(8x^2)]^2}}$$

$$\displaystyle{y=\ln\sqrt{\frac{x^2+1}{x+3}}}$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{y=\ln\sqrt{\frac{x^2+1}{x+3}}}$$.

$$\displaystyle{y'=\frac{x}{x^2+1}-\frac{1}{2(x+3)}}$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{y=\ln\sqrt{\frac{x^2+1}{x+3}}}$$.

Solution

### 1939 solution video

video by PatrickJMT

$$\displaystyle{y'=\frac{x}{x^2+1}-\frac{1}{2(x+3)}}$$

$$\displaystyle{ f(x)=\left[ \frac{x+4}{\sqrt{x^2+1}} \right]^3 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ f(x)=\left[ \frac{x+4}{\sqrt{x^2+1}} \right]^3 }$$.

Solution

### 989 solution video

video by PatrickJMT

$$\displaystyle{ y=x \sin(1/x) + \sqrt[4]{(1-3x)^2 + x^5} }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y=x \sin(1/x) + \sqrt[4]{(1-3x)^2 + x^5} }$$.

Solution

### 990 solution video

video by PatrickJMT

$$\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }$$

Problem Statement

Use the chain rule to calculate the derivative of $$\displaystyle{ y=\left[ \tan \left( \sin \left( \sqrt{x^2+8x}\right) \right) \right]^5 }$$.

Solution

### 988 solution video

video by PatrickJMT