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AP Calculus AB : Derivatives

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Example Questions

Example Question #5 : Equations Involving Derivatives

Practicing the chain rule level 2 C!

Find the derivative of the function

$y(x)=\frac{-1}{2^{x}-4}$

Possible Answers:

$y'(x)=\frac{-ln(2)}{2^x-4}$

$y'(x)=\frac{-1}{ln(2)(2^x-4)^2}$

$y'(x)=\frac{2^x*ln(2)}{(2^x-4)^2}$

$y'(x)=\frac{-ln(2)(2^x-4)}{(2^x-4)^2}$

$y'(x)=\frac{2x}{2^x-4}$

Correct answer:

$y'(x)=\frac{2^x*ln(2)}{(2^x-4)^2}$

Explanation:

To understand why the answer is

$y'(x)=\frac{2^x*ln(2)}{(2^x-4)^2}$

you must understand that the derivative of

$y(x)=\frac{-1}{x}$

is actually

$y'(x)=\frac{1}{x^2}$ .

Next, you must understand that the derivative of

y(x)=2^x-4

is actually

y'(x)=2^x*ln(2) .

$y(x)=\frac{-1}{2^x-4}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x}$ and g(x)=2^x-4 .

y(x) in terms of f(x) and g(x) is actually f(g(x)) which means

$f(g(x))=\frac{-1}{2^x-4}$ since in $\frac{-1}{x}$ is substituted with 2^x-4 .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable, one step at a time and multiply the results together.

the derivative of

y(x)=f(g(x))

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

y(x)'=f'(...)

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

y(x)'=f'(g(x))...

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

y(x)'=f'(g(x))*g'(x)

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$y'(x)=\frac{2^x*ln(2)}{(2^x-4)^2}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner function while keeping the next inner functions the the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

y=f(g(h(x)))

then it's derivative would be

y'=f'(g(h(x)))*g'(h(x))*h'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

Chain rule table

Report an Error

Example Question #3 : Equations Involving Derivatives

Practicing the chain rule level 2 D!

Find the derivative of the function

y(x)=ln(arccos(x))

Possible Answers:

$y'(x)=\frac{-1}{arccos(x)}$

$y'(x)=\sqrt{1-x^2}$

$y'(x)=\frac{\sqrt{1-x^2}}{arccos(x)}$

$y'(x)=\frac{-1}{\sqrt{1-x^2}*arccos(x)}$

$y'(x)=\frac{ln(arccos(x))}{\sqrt{1-x^2}}$

Correct answer:

$y'(x)=\frac{-1}{\sqrt{1-x^2}*arccos(x)}$

Explanation:

To understand why the answer is

$y'(x)=\frac{-1}{\sqrt{1-x^2}*arccos(x)}$ ,

you must understand that the derivative of

y(x)=ln(x)

is actually

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

Next, you must understand that the derivative of

y(x)=arccos(x)

is actually

$y'(x)=\frac{-1}{\sqrt{1-x^2}}$ .

y(x)=ln(arccos(x)) can be treated as a composition of the functions

$f(x)=\frac{1}{x}$ and g(x)=arccos(x) .

y(x) in terms of f(x) and g(x) is actually f(g(x)) which means

f(g(x))=ln(arccos(x)) since in ln(x) is substituted with .

the derivative of

y(x)=f(g(x))

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

y(x)'=f'(...)

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

y(x)'=f'(g(x))...

Step 3: Differentiate the the next function g(x) but multiply it by f'(g(x))

y(x)'=f'(g(x))*g'(x)

So, substitute f'(x), g(x) and g'(x) for the expressions you found before:

$y'(x)=\frac{-1}{\sqrt{1-x^2}*arccos(x)}$

And now you have found the correct answer.

-------------------------------------------------------------------------------------------

If we have

y=f(g(h(x)))

then it's derivative would be

y'=f'(g(h(x)))*g'(h(x))*h'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in these equations

Chain rule table

Report an Error

Example Question #4 : Equations Involving Derivatives

Practicing the chain rule level 3 A!

Find the derivative of the function

$y(x)=ln(\sqrt{sin(x^2)})$

Possible Answers:

y'(x)=tan(x^2)

y'(x)=2x*sin(x^2)*cos(x)

y'(x)=x*cot(x^2)

$y'(x)=\frac{1}{2}ln(sin(x^{2}))*cos(x^2)$

$y'(x)=\frac{x}{cos(x^2)}$

Correct answer:

y'(x)=x*cot(x^2)

Explanation:

To understand why the answer is

y(x)'=x*cot(x^2) ,

first remember that ln(x^a)=a*ln(x) .

Then, you must understand that the derivative of

y(x)=ln(x)

is actually

y'(x)=1/x .

Next, you must understand that the derivative of

y(x)=sin(x)

is actually

y'(x)=cos(x) .

And finally, you must understand that the derivative of

y(x)=x^2

is actually

y'(x)=2x .

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

f(x)=ln(x) , $g(x)=\sqrt{sin(x)}$ and h(x)=x^2 .

The reason why we did not define $g(x)=\sqrt{x}$ , h(x)=sin(x) and m(x)=x^2

is because of the property ln(x^a)=a*ln(x) mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}*ln(g(h(x)))$

y(x) in terms of f(x) , g(x) and h(x) is actually f(g(h(x))) which means

$f(g(h(x)))=\frac{1}{2}*ln(sin(x^2))$ since in ln(x) is substituted with sin(x) and in sin(x) can be substituted with x^2 .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

y(x)=f(g(h(x)))

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

y'(x)=f'(...)

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

y'(x)=f'(g(...))...

Step 3: Look at the next function h(x), keep it inside the other function g(x)

y'(x)=f'(g(h(x)))

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

y'(x)=f'(g(h(x)))*g'(h(x))...

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))*g'(h(x))

f'(g(h(x)))*g'(h(x))*h'(x)

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}*cos(x^2)*2x$ which is x*cot(x^2)

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to find the derivative of. This also applies to more complicated functions. For instance,

If we have

y=f(g(h(m(x))))

then it's derivative would be

y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Chain rule table

Report an Error

Example Question #5 : Equations Involving Derivatives

Practicing the chain rule level 3 B!

Find the derivative of the function

q(z)=-sin(cos(tan(z)))

Possible Answers:

$-cos(sin(tan(z)))*cos(tan(z))*sec^{2}(z)$

-sin(cos(tan(z)))*cos(sin(sec(z)))

cos(cos(z))*sin(tan(z))*sec(z)

$sin(cos(sec^{2}(z)))*cos(sec^{2}(z))*sec^{2}(z)$

$cos(cos(tan(z)))*sin(tan(z))*sec^{2}(z)$

Correct answer:

$cos(cos(tan(z)))*sin(tan(z))*sec^{2}(z)$

Explanation:

To understand why the answer is

$q(z)=cos(cos(tan(z)))*sin(tan(z))*sec^{2}(z)$ ,

you must understand that the derivative of

q(z)=sin(z)

is actually

q'(z)=cos(z) .

Next, you must understand that the derivative of

q(z)=cos(z)

is actually

q'(z)=-sin(z) .

And finally, you must understand that the derivative of

q(z)=tan(z)

is actually

$q'(z)=sec^{2}(z)$ .

q(z)=-sin(cos(tan(z))) can be treated as a composition of the functions

f(z)=-sin(z) , g(z)=cos(z) and h(z)=tan(z) .

q(z) in terms of f(z) , g(z) and h(z) is actually f(g(h(z))) which means

f(g(h(z)))=-sin(cos(tan(z))) since in -sin(z) is substituted with cos(z) and in cos(z) can be substituted with tan(z) .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

q(z)=f(g(h(z)))

is...

Step 1: Only look at the outermost function f(z) first, then differentiate it

q'(z)=f'(...)

Step 2: Look at the next function g(z), keep it inside the other function f'(z)

q'(z)=f'(g(...))...

Step 3: Look at the next function h(z), keep it inside the other function g(z)

q'(z)=f'(g(h(z)))

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

q'(z)=f'(g(h(z)))*g'(h(z))...

Step 5: Differentiate the next function h(z) but multiply it by the factors f'(g(h(z)))*g'(h(z))

f'(g(h(z)))*g'(h(z))*h'(z)

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(z), g(z), g'(z), h(z) and h'(z) for the expressions you found before:

$cos(cos(tan(z)))*sin(tan(z)))*sec^{2}(z)$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

To sum it up, you multiply the derivative of the outermost functions by the derivative if the inner functions while keeping the next inner functions the same in every step until you are out of functions to differentiate. This also applies to more complicated functions. For instance,

If we have

y=f(g(h(m(x))))

then it's derivative would be

y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Chain rule table

Report an Error

Example Question #71 : Derivative As A Function

Practicing the chain rule level 1 E!

Find the derivative of the function

y(x)=ln(-4x+3)

Possible Answers:

$y'(x)=\frac{-4}{3-4x}$

$y'(x)=\frac{ln(-4x-3)}{4x-3}$

$y'(x)=\frac{4}{4x-3}$

$y'(x)=\frac{4x-3}{ln(3-4x)}$

$y'(x)=\frac{-4}{x}$

Correct answer:

$y'(x)=\frac{4}{4x-3}$

Explanation:

To understand why the answer is

$y(x)'=\frac{4}{4x-3}$ ,

you must understand that the derivative of

y(x)=ln(x)

is actually

y'(x)=1/x .

Next, you must understand that the derivative of

y(x)=3-4x

is actually

y'(x)=-4 .

And finally, you must understand that the derivative of

y(x)=x^2

is actually

y'(x)=2x .

$y(x)=ln(\sqrt{sin(x^2)})$ can be treated as a composition of the functions

f(x)=ln(x) , $g(x)=\sqrt{sin(x)}$ and h(x)=x^2 .

The reason why we did not define $g(x)=\sqrt{x}$ , h(x)=sin(x) and m(x)=x^2

is because of the property ln(x^a)=a*ln(x) mentioned before,

turning

$ln(\sqrt{g(h(x))})$

into

$\frac{1}{2}*ln(g(h(x)))$

y(x) in terms of f(x) , g(x) and h(x) is actually f(g(h(x))) which means

$f(g(h(x)))=\frac{1}{2}*ln(sin(x^2))$ since in ln(x) is substituted with sin(x) and in sin(x) can be substituted with x^2 .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

y(x)=f(g(h(x)))

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

y'(x)=f'(...)

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

y'(x)=f'(g(...))...

Step 3: Look at the next function h(x), keep it inside the other function g(x)

y'(x)=f'(g(h(x)))

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

y'(x)=f'(g(h(x)))*g'(h(x))...

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))*g'(h(x))

f'(g(h(x)))*g'(h(x))*h'(x)

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\frac{1}{2}\frac{1}{sin(x^2)}*cos(x^2)*2x$ which is x*cot(x^2)

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

If we have

y=f(g(h(m(x))))

then it's derivative would be

y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Chain rule table

Report an Error

Example Question #551 : Derivatives

Practicing the chain rule level 3 C!

Find the derivative of the function

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

Possible Answers:

$y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

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

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

$y'(x)=\frac{-x-2}{-2x^22ln(\frac{1}{(x^2+3)})}$

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

Correct answer:

$y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

Explanation:

To understand why the answer is

$y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$ ,

Then, you must understand that the derivative of

$\large y(x)=\frac{1}{x}$

is actually

$\large y'(x)=\frac{-1}{x^2}$ .

Next, you must understand that the derivative of

$\large y(x)=ln(x)$

is actually

$\large y'(x)=\frac{1}{x}$ .

And finally, you must understand that the derivative of

$\frac{1}{(2x+3)^2}$

is actually

$y'(x)=\frac{-4x}{(x^2+3)^3}$ .

$y(x)=\frac{1}{ln((x^2+3)^{-2})}$ can be treated as a composition of the functions

$f(x)=\frac{1}{x+a}$ , g(x)=ln(x+a) and h(x)=x^2+a .

y(x) in terms of f(x) , g(x) and h(x) is actually f(g(h(x))) which means

$f(g(h(x)))=\frac{1}{ln((x^2+3)^{-2})}$ since in $\frac{1}{x+a}$ is substituted with ln(x) and in ln(x) can be substituted with $\frac{1}{(x^2+3)^2}$ .

This means in order to differentiate the equation, you must use the chain rule. The chain rule tells us that in order to differentiate a composition function, you must differentiate all of the composite functions in the reverse order that they are applied to the variable one step at a time and multiply the results together.

the derivative of

y(x)=f(g(h(x)))

is...

Step 1: Only look at the outermost function f(x) first, then differentiate it

y'(x)=f'(...)

Step 2: Look at the next function g(x), keep it inside the other function f'(x).

y'(x)=f'(g(...))...

Step 3: Look at the next function h(x), keep it inside the other function g(x)

y'(x)=f'(g(h(x)))

Step 4: Differentiate the the next function g(x) while keeping h(x) inside of g(x) and g'(x). But, multiply g'(h(x)) by f'(g(h(x)))

y'(x)=f'(g(h(x)))*g'(h(x))...

Step 5: Differentiate the next function h(x) but multiply it by the factors f'(g(h(x)))*g'(h(x))

f'(g(h(x)))*g'(h(x))*h'(x)

Since you are out of composite functions to differentiate, stop here.

Now, substitute f'(x), g(x), g'(x), h(x) and h'(x) for the expressions you found before:

$\large \large y'(x)=\frac{4x}{(x^2+3)*ln^2(\frac{1}{(x^2+3)^2})}$

and now you have found the correct answer.

-------------------------------------------------------------------------------------------

If we have

y=f(g(h(m(x))))

then it's derivative would be

y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)

Notice how the factors become less complicated as you differentiate it or as you look from left to right.

If you are unsure of the pattern, look at the pattern in the table below.

Chain rule table

Report an Error

Example Question #552 : Derivatives

Find the eqation of the line tangent to the graph of $y=x+x^{1/2}$ at point (4,6) .

Possible Answers:

$y'=1+\frac{1}{2(x^{1/2})}$

$y=1+\frac{1}{2}x^{1/2}$

$y=\frac{1}{2x^{1/2}}$

$y=1+\frac{1}{2(6^{1/2})}$

$y=\frac{5}{4}x+1$

Correct answer:

$y=\frac{5}{4}x+1$

Explanation:

$y=x+x^{1/2}$

$y'=1+\frac{1}{2}x^{-1/2}$

$y'=1+\frac{1}{2(x^{1/2})}$

Now we plug in x=4 .

$y'=1+\frac{1}{2(4^{1/2})} = 1+\frac{1}{4} = \frac{5}{4}$ this is the slope.

Now use the point slope formula to find the tangent line.

$y-6=\frac{5}{4}(x-4)$

$y-6=\frac{5}{4}x-5$

$y=\frac{5}{4}x+1$

Report an Error

Example Question #21 : Applications Of Derivatives

Find the rate of change of y if $y=5x^{-3}+3e^{x}$

Possible Answers:

$y'=6x^{3}+e^{3x}$

$y'=-15x^{-4}+3e^{x}$

$y'=-6x^{-2}+3e^{3x}$

$y'=-15x^{-3}+4e^{x}$

$y'=-3x^{-4}+3e^{x}$

Correct answer:

$y'=-15x^{-4}+3e^{x}$

Explanation:

The rate of change of y is also the derivative of y.

Differentiate the function given.

You should get $y'=-15x^{-4}+3e^{x}$

Report an Error

Example Question #22 : Applications Of Derivatives

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

p(t)=-365sin(t)+12t^3-3t^2+164

Possible Answers:

p'(t)=365cos(t)+36t^3-6t

p'(t)=-365cos(t)+36t^3-6t

p'(t)=-365+cos(t)+36t^3-6t

p'(t)=-365cos(t)+36t^3-6t+164

Correct answer:

p'(t)=-365cos(t)+36t^3-6t

Explanation:

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

p(t)=-365sin(t)+12t^3-3t^2+164

Begin by recalling that velocity is the first derivative of position. So all we need to do is find the first derivative of our position function.

Recall that the derivative of sine is cosine, and that the derivative of polynomials can be found by multiplying each term by its exponent, and decreasing the exponent by 1.

Starting with:

p(t)=-365sin(t)+12t^3-3t^2+164

We get:

p'(t)=-365cos(t)+36t^3-6t

Report an Error

Example Question #23 : Applications Of Derivatives

Given j(k), find the rate of change when k=5.

j(k)=-3k^3+4k+16

Possible Answers:

125

Correct answer:

Explanation:

Given j(k), find the rate of change when k=5

j(k)=-3k^3+4k+16

Let's begin by realizing that a rate of change refers to a derivative.

So, we need to find the derivative of j(k)

j(k)=-3k^3+4k+16

We find this by multiplying each term by the exponent, and decreasing the exponent by 1

j(k)=-3k^3+4k+16

j'(k)=-9k^2+4

Next, plug in 5 to find our answer:

j'(5)=-9(5)^2+4=-9*25+4=-225+4=-221

So, our rate of change is -221.

Report an Error

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #5 : Equations Involving Derivatives

Example Question #3 : Equations Involving Derivatives

Example Question #4 : Equations Involving Derivatives

Example Question #5 : Equations Involving Derivatives

Example Question #71 : Derivative As A Function

Example Question #551 : Derivatives

Example Question #552 : Derivatives

Example Question #21 : Applications Of Derivatives

Example Question #22 : Applications Of Derivatives

Example Question #23 : Applications Of Derivatives

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