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Calculus AB : Differentiating Functions

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Example Question #15 : Implicit Differentiation And Chain Rule

Find the derivative of the following function:

f(x)=ln(3x) .

Possible Answers:

ln(3x)

$\frac{1}{3x}$

$\frac{3}{x}$

3ln(3x)

$\frac{1}{x}$

Correct answer:

$\frac{1}{x}$

Explanation:

This is a chain rule derivative. We must first differentiate the natural log function, leaving the inner function as is. Recall:

$ln(x)'=\frac{1}{x}$

Now, we must replace this with our function, and multiply that by the derivative of the inner function:

$ln(3x)'=\frac{1}{3x}(3)=\frac{3}{3x}=\frac{1}{x}.$

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Example Question #16 : Implicit Differentiation And Chain Rule

Find the derivative of b(x)=ln(cos(x)) .

Possible Answers:

b'(x)=-cot(x)

b'(x)=tan(x)

b'(x)=-tan(x)

$b'(x)=-\frac{1}{sin(x)}$

b'(x)=-sin(x)

Correct answer:

b'(x)=-tan(x)

Explanation:

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 -sin(x) .

Multiplying these two together gives -tan(x) .

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Example Question #17 : Implicit Differentiation And Chain Rule

If $f(x)=\ln(e^{2x^2})$ , calculate f'(4)

Possible Answers:

$2\ln2$

$\ln2$

$e^{16}$

Correct answer:

Explanation:

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, f'(4) = 4(4)=16 .

Notice that we could have also simplified f(x) first by cancelling the natural log and the exponential function leaving us with just f(x) =2x^2 , thereby avoiding the chain rule altogether.

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Example Question #18 : Implicit Differentiation And Chain Rule

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

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

Possible Answers:

$\frac{dy}{dx}=\frac{\sin(x)-1}{2xy +\cos(x)}$

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

$\frac{dy}{dx}=\frac{\sin(x)-3}{2x +\cos(x)}$

$\frac{dy}{dx}=\frac{2y\sin(x)-3y^2}{2y -\cos(x)}$

$\frac{dy}{dx}=\frac{2\sin(x)+3y^2}{2y -x\sin(x)}$

Correct answer:

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

Explanation:

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

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,

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

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

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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 $\frac{d}{dx}(y^2)=2y\frac{dy}{dx}$ .

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

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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Example Question #11 : Implicit Differentiation And Chain Rule

Find the derivative of f(x)=(1+3x^2)^5 .

Possible Answers:

30x(1+3x^2)^5

30x(1+3x^2)^4

30x(1+3x^2)

5(1+3x^2)^4

30(1+3x^2)^4

Correct answer:

30x(1+3x^2)^4

Explanation:

This is a chain rule derivative. We must first start by taking the derivative of the outermost function. Here, that is a function raised to the fifth power. We need to take that derivative (using the power rule). Then, we multiply by the derivative of the innermost function:

$f'(x)=5(1+3x^2)^{5-1}*6x=30x(1+3x^2)^4.$

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Example Question #11 : Implicit Differentiation And Chain Rule

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

Possible Answers:

$y'=e^{sinx^3}*cosx^3*3x^2$

$y'=e^{cosx^3}$

$y'=e^{sinx^3}*cosx^3$

$y'=sinx^3(e^{sinx^3-1})$

$y'=e^{cosx^3}*3x^2$

Correct answer:

$y'=e^{sinx^3}*cosx^3*3x^2$

Explanation:

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 cosx^3 * 3x^2 . So, the final answer is $y'=e^{sinx^3}*cosx^3*3x^2$ .

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Calculus AB : Differentiating Functions

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #15 : Implicit Differentiation And Chain Rule

Example Question #16 : Implicit Differentiation And Chain Rule

Example Question #17 : Implicit Differentiation And Chain Rule

Example Question #18 : Implicit Differentiation And Chain Rule

Example Question #11 : Implicit Differentiation And Chain Rule

Example Question #11 : Implicit Differentiation And Chain Rule

All Calculus AB Resources

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