Chain rule and implicit differentiation - AP Calculus AB

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Question

$f(x)=sin(x^{2}})$ . Find .

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Answer

To take the derivative, you must first take the derivative of the outside function, which is sine. However, the , or the angle of the function, remains the same until we take its derivative later. The derivative of sinx is cosx, so you the first part of will be . Next, take the derivative of the inside function, . Its derivative is , so by the chain rule, we multiply the derivatives of the inside and outside functions together to get .

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Question

. Using the chain rule for derivatives, find .

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Answer

By the chain rule, we must first take the derivative of the outside function by bringing the power down front and reducing the power by one. When we do this, we do not change the function that is in the parentheses, or the inside function. That means that the first part of will be . Next, we must take the derivative of the inside function. Its derivative is . The chain rule says we must multiply the derivative of the outside function by the derivative of the inside function, so the final answer is .

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Question

$f(x)=6^(^5^x^{^{2}-7x}^)$ . Find the derivative.

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Answer

When the function is a constant to the power of a function of x, the first step in chain rule is to rewrite f(x). So, the first factor of f(x) will be $(6^{5x^2-7x})$ . Next, we have to take the derivative of the function that is the exponent, or . Its derivative is 10x-7, so that is the next factor of our derivative. Last, when a constant is the base of an exponential function, we must always take the natural log of that number in our derivative. So, our final factor will be . Thus, the derivative of the entire function will be all these factors multiplied together: $f'(x)=6^{5x^2-7x}(10x-7)ln6}$ .

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Question

Find the derivative of:

$h(t)=\ln(\ln t)$

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Answer

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\ln(t)$

and

$f(t)=\ln(t)$ .

Since

$g'(t)=\frac{1}{t}$

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=\frac{1}{f(t)}\cdot\frac{1}{t}=\frac{1}{t\ln t}$

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Question

Determine the derivative of $f(x)=2\tan ^2(x^2)$

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Answer

This is a pure problem on understanding how chain rules work for derivatives.

First thing we need to remember is that the derivative of $\tan(x)$ is $\sec^2(x)$ .

When we are taking the derivative of $f(x)=2\tan ^2(x^2)$ , we can first pull out the 2 in the front and we treat $\tan^2(x^2)$ as $[\tan(x^2)]^2$ .

This way, the derivative will become $22\tan(x^2)\frac{\mathrm{d} tan(x^2)}{\mathrm{d} x}$ ,

which is $4\tan(x^2)(2x\sec(x^2))$ .

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Question

Find the derivative of the function: $y=e^{sinx^3}}$ .

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Answer

Whenever we have an exponential function with , the first term of our derivative will be that term repeated, without changing anything. So, the first factor of the derivative will be $e^{sinx^3}$ . Next, we use chain rule to take the derivative of the exponent. Its derivative is . So, the final answer is $y'=e^{sinx^3}cosx^33x^2$ .

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Question

Find the derivative of the exponential function, $y=e^{sec3\theta }$ .

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Answer

To take the derivative of any exponential function, we repeat the exponential function in the derivative. So, the first factor of the derivative will be $e^{sec3\theta}$ . Next, we have to take the derivative of the exponent using chain rule. The derivative of the trigonometric function secx is secxtanx, so in terms of this problem its derivative is $sec(3\theta)tan(3\theta)$ . Since the angle has a scalar of 3, we must also multiply the entire derivative by 3. So, the answer is $y'=3e^{sec3\theta}*sec(3\theta)tan(3\theta)$ .

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Question

Find the derivative of $f(x)=10^{2\sqrt{x}}$ .

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Answer

To find the derivative, we can first rewrite the function to make it easier to take the chain rule. Rewrite $10^{2\sqrt{x}}$ as $10^{2x^{\frac{1}{2}}}$ . Now, like in any exponential function, the first factor of the derivative is the original exponential function. So, the first factor of f'(x) will be $10^{2\sqrt{x}}$ . Next, by the chain rule for derivatives, we must take the derivative of the exponent, which is why we rewrote the exponent in a way that is easier to take the derivative of. So, the derivative of the exponent is $x^{-{\frac{1}{2}}}$ , because the 1/2 and the 2 cancel when we bring the power down front, and the exponent of 1/2 minus 1 becomes negative 1/2. The last factor of the derivative is because in every derivative of an exponential function where the base is a number, we must multiply by the natural log of that base. So, once you multiply all these factors together, the final answer is $f'(x)=10^{2\sqrt{x}}ln10x^{-{\frac{1}{2}}}$

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Question

If , find the derivative through implicit differentiation.

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Answer

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get The next step is to solve for y', so we put all terms containing y' on the left side of the equation: . Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: To get y' alone, divide both sides by to get $y'=\frac{6y-3x^2}{3y^2-6x}$ . To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them. So, the final answer is $y'=\frac{2y-x^2}{y^2-2x}$ .

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Question

If , find .

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Answer

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get The next step is to solve for y', so we put all terms containing y' on the left side of the equation and factor out a common y': . To get y' alone, divide both sides by to get $y'=\frac{-y^2sinx-cos(x+y)}{cos(x+y)-2ycosx}$ .

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Question

Find the derivative of the function of the circle

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Answer

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get The next step is to solve for y', so we put all terms containing y' on the left side of the equation: . To get y' alone, divide both sides by to get $y'=\frac{-2x}{2y}$ . To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them. So, the final answer is $\frac{-x}{y}$ .

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Question

Find the derivative of the function using implicit differentiation.

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Answer

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get . The next step is to solve for y', so we put all terms containing y' on the left side of the equation: . To get y' alone, divide both sides by -3 to get $y'=\frac{-12x^2}{-3}$ . To simplify even further, we can factor a -2 out of the numerator and denominator and cancel them. So, the final answer is .

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Question

Find the derivative of the function .

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Answer

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x. Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'. So, when we take the derivative of each term, we get . The next step is to solve for y', so we put all terms containing y' on the left side of the equation: . Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: . To get y' alone, divide both sides by to get a final answer of $y'=\frac{3-2xy}{x^2+6y^2-2}$ .

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Question

Find the derivative of the function .

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Answer

Before we take the derivative of the logarithmic function, we can make it easier for ourselves by simplifying the equation to . We can bring the exponent of 6 down in front of the natural log of x due to properties of logarithms. Next, take the derivative of each term in terms of x. Don't forget to multiply by y' each time you take the derivative of a term containing y! When we do this, we should get $(\frac{1}{y})y'=6(\frac{1}{x})$ because the derivative of lnx is 1/x. Next, solve for y' by multiplying both sides by y to get the final answer of $y'=\frac{6y}{x}$ .

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Question

Differentiate,

$y = \sin(\sin (x))$

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Answer

$y = \sin(\sin x)$ (1)

An easier way to think about this:

Because is a function of a function, we must apply the chain rule. This can be confusing at times especially for function like equation (1). The differentiation is easier to follow if you use a substitution for the inner function,

Let,

$u = \sin(x)$ (2)

So now equation (1) is simply,

$y =\sin (u)$ (3)

Note that is a function of . We must apply the chain rule to find $\frac{dy}{dx}$ ,

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ (4)

To find the derivatives on the right side of equation (4), differentiate equation (3) with respect to , then Differentiate equation (2) with respect to .

$\frac{dy}{du}=\cos(u)$ $\frac{du}{dx}=\cos(x)$

Substitute into equation (4),

$\frac{dy}{dx}=\cos(u)\cos(x)$ (5)

Now use $u = \sin(x)$ to write equation (5) in terms of alone:

$\frac{dy}{dx} =\cos(\sin(x))\times\cos(x)$

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Question

Use implicit differentiation to find $\frac{dy}{dx}$ is terms of and for,

$xy^2=y\sin(x)+1$

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Answer

$xy^2=y\sin(x)+1$

To differentiate the equation above, start by applying the derivative operation to both sides,

$\frac{d}{dx}(xy^2)=\frac{d}{dx}[y\sin(x)]+\underset{=0}{ \underbrace{\frac{d}{dx}(1)}}$

Both sides will require the product rule to differentiate,

$y^2\frac{d}{dx}(x) +x\frac{d}{dx}(y^2) =\sin(x)\frac{d}{dx}(y)+y\frac{d}{dx}\sin(x)$

$y^2\underset{=1}{ \underbrace{\frac{d}{dx}(x)}} +x\left(2y\frac{dy}{dx} \right ) =\sin(x)\frac{dy}{dx}+y\cos(x)$ Common Mistake

A common mistake in the previous step would be to conclude that $\frac{d}{dx}(y^2)=2y$ instead of $2y\frac{dy}{dx}$ . The former is not correct; if we were looking for the derivative with respect to , then $\frac{d}{dy}(y^2)$ would in fact be . But we are not differentiating with respect to , we're looking for the derivative with respect to .

We are assuming that is a function of , so we must apply the chain rule by differentiating with respect to and multiplying by the derivative of with respect to to obtain $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ . $y^2+2xy\frac{dy}{dx} =\sin(x)\frac{dy}{dx}+y\cos(x)$

Collect terms with a derivative onto one side of the equation, factor out the derivative, and divide out to solve for the derivative $\frac{dy}{dx}$ ,

$2xy\frac{dy}{dx} -\sin(x)\frac{dy}{dx}=y\cos(x)-y^2$

$\frac{dy}{dx}[2xy -\sin(x)]=y\cos(x)-y^2$

$\frac{dy}{dx}=\frac{y\cos(x)-y^2}{2xy -\sin(x)}=\frac{y(\cos x-y)}{2xy -\sin (x)}$

Therefore,

$\frac{dy}{dx}=\frac{y(\cos x-y)}{2xy -\sin x}$

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Question

Find given

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Answer

Here we use the chain rule: $\frac{\mathrm{d} }{\mathrm{d} x}ln(g(x))=\frac{1}{g(x)}\cdot g'(x)$

Let

Then

And $f'(x)=\frac{1}{x^3+x^2}\cdot (3x^2+2x)=\frac{x(3x+2)}{x(x^2+x)}=\frac{3x+2}{x^2+x}$

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Question

If $f(x)=\ln(e^{2x^2})$ , calculate

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Answer

Using the chain rule, we have

$f'(x) = \frac{1}{e^{2x^2}}\times (e^{2x^2})'$

$= \frac{1}{e^{2x^2}}\times e^{2x^2} \times 4x = 4x$ .

Hence, .

Notice that we could have also simplified first by cancelling the natural log and the exponential function leaving us with just , thereby avoiding the chain rule altogether.

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Question

Use the chain rule to find the derivative of the function $y=(3x+1)^{2}$

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Answer

First, differentiate the outside of the parenthesis, keeping what is inside the same.

You should get .

Next, differentiate the inside of the parenthesis.

You should get .

Multiply these two to get the final derivative .

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Question

Find the derivative of .

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Answer

Use chain rule to solve this. First, take the derivative of what is outside of the parenthesis.

You should get $\frac{1}{cos(x)}$ .

Next, take the derivative of what is inside the parenthesis.

You should get .

Multiplying these two together gives .

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