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AP Calculus AB : Derivative as a function

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Example Questions

Example Question #71 : Derivative As A Function

Practicing the chain rule level 3 B!

Find the derivative of the function

$\displaystyle q(z)=-sin(cos(tan(z)))$

Possible Answers:

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

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

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

$\displaystyle cos(cos(z))*sin(tan(z))*sec(z)$

$\displaystyle -sin(cos(tan(z)))*cos(sin(sec(z)))$

Correct answer:

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

Explanation:

To understand why the answer is

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

you must understand that the derivative of

$\displaystyle q(z)=sin(z)$

is actually

$\displaystyle q'(z)=cos(z)$.

Next, you must understand that the derivative of

$\displaystyle q(z)=cos(z)$

is actually

$\displaystyle q'(z)=-sin(z)$.

And finally, you must understand that the derivative of

$\displaystyle q(z)=tan(z)$

is actually

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

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

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

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

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

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

$\displaystyle q(z)=f(g(h(z)))$

is...

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

$\displaystyle q'(z)=f'(...)$

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

$\displaystyle q'(z)=f'(g(...))...$

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

$\displaystyle q'(z)=f'(g(h(z)))$

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)))

$\displaystyle q'(z)=f'(g(h(z)))*g'(h(z))...$

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

$\displaystyle f'(g(h(z)))*g'(h(z))*h'(z)$

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:

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

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

$\displaystyle y=f(g(h(m(x))))$

then it's derivative would be

$\displaystyle y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)$

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

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Example Question #72 : Derivative As A Function

Practicing the chain rule level 1 E!

Find the derivative of the function

$\displaystyle y(x)=ln(-4x+3)$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To understand why the answer is

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

you must understand that the derivative of

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

is actually

$\displaystyle y'(x)=1/x$.

Next, you must understand that the derivative of

$\displaystyle y(x)=3-4x$

is actually

$\displaystyle y'(x)=-4$.

And finally, you must understand that the derivative of

$\displaystyle y(x)=x^2$

is actually

$\displaystyle y'(x)=2x$.

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

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

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

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

turning

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

into

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

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

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

the derivative of

$\displaystyle y(x)=f(g(h(x)))$

is...

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

$\displaystyle y'(x)=f'(...)$

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

$\displaystyle y'(x)=f'(g(...))...$

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

$\displaystyle y'(x)=f'(g(h(x)))$

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)))

$\displaystyle y'(x)=f'(g(h(x)))*g'(h(x))...$

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

$\displaystyle f'(g(h(x)))*g'(h(x))*h'(x)$

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:

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

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

$\displaystyle y=f(g(h(m(x))))$

then it's derivative would be

$\displaystyle y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)$

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

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Example Question #73 : Derivative As A Function

Practicing the chain rule level 3 C!

Find the derivative of the function

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To understand why the answer is

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

Then, you must understand that the derivative of

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

is actually

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

Next, you must understand that the derivative of

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

is actually

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

And finally, you must understand that the derivative of

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

is actually

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

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

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

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

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

the derivative of

$\displaystyle y(x)=f(g(h(x)))$

is...

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

$\displaystyle y'(x)=f'(...)$

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

$\displaystyle y'(x)=f'(g(...))...$

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

$\displaystyle y'(x)=f'(g(h(x)))$

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)))

$\displaystyle y'(x)=f'(g(h(x)))*g'(h(x))...$

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

$\displaystyle f'(g(h(x)))*g'(h(x))*h'(x)$

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:

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

and now you have found the correct answer.

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

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

$\displaystyle y=f(g(h(m(x))))$

then it's derivative would be

$\displaystyle y'=f'(g(h(m(x))))*g'(h(m(x)))*h'(m(x))*m'(x)$

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 #11 : Equations Involving Derivatives

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Now we plug in x=4 $\displaystyle x=4$ .

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

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

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

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

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

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Example Question #681 : Derivatives

At the point x=2 $\displaystyle x=2$, is the function $\displaystyle -8x^2+15$ increasing or decreasing, concave or convex?

Possible Answers:

The function is undefined at that point

Increasing, convex

Decreasing, concave

Decreasing, convex

Increasing, concave

Correct answer:

Decreasing, convex

Explanation:

First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat $\displaystyle 15$ as 15x^0 $\displaystyle 15x^0$ since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ $\displaystyle (0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

Plug in our given point for $\displaystyle x$. If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

y=-16x $\displaystyle y=-16x$

$\displaystyle y=-16(2)$

y=-32 $\displaystyle y=-32$

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using -16x $\displaystyle -16x$ as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=1*-16x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16x^{0}$

$\frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for $\displaystyle x$; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

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Example Question #1 : Finding Regions Of Concavity And Convexity

When x=3 $\displaystyle x=3$, what is the concavity of the graph of $2x^4+\frac{1}{2}x$ $\displaystyle 2x^4+\frac{1}{2}x$?

Possible Answers:

There is insufficient data to solve.

Increasing, concave

Increasing, convex

Decreasing, convex

Decreasing, concave

Correct answer:

Increasing, convex

Explanation:

To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{4-1})+(1*\frac{1}{2}x^{1-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{4-1})+(1*\frac{1}{2}x^{1-1})$

Simplify:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{3})+(\frac{1}{2}x^{0})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(4*2x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}*1)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, x=3 $\displaystyle x=3$, to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

$y=8(3)^{3}+\frac{1}{2}$ $\displaystyle y=8(3)^{3}+\frac{1}{2}$

y=216.5 $\displaystyle y=216.5$

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, x=3 $\displaystyle x=3$, and check if it is positive or negative.

We're going to treat $\frac{1}{2}$ $\displaystyle \frac{1}{2}$ as $\frac{1}{2}x^0$ $\displaystyle \frac{1}{2}x^0$ since anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{3-1})+(0*\frac{1}{2}x^{0-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{3-1})+(0*\frac{1}{2}x^{0-1})$

Notice that $(0*\frac{1}{2}x^{0-1})=0$ $\displaystyle (0*\frac{1}{2}x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{2})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=(3*8x^{2})$

$\frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(8x^{3}+\frac{1}{2})=24x^{2}$

Plug in our given value:

$\displaystyle y=24(3)^2$

y=216 $\displaystyle y=216$

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

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Example Question #1 : Finding Regions Of Concavity And Convexity

At the point x=-5 $\displaystyle x=-5$, is $\displaystyle 5x^2+12x$ increasing or decreasing, and is it concave or convex?

Possible Answers:

Decreasing, concave

Decreasing, convex

Increasing, convex

Increasing, concave

The graph is undefined at point x=-5 $\displaystyle x=-5$

Correct answer:

Decreasing, convex

Explanation:

To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

$\displaystyle y=10x+12$

$\displaystyle y=10(-5)+12$

$\displaystyle y=-50+12$

y=-38 $\displaystyle y=-38$

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using 10x+12 $\displaystyle 10x+12$ as our expression.

We're going to treat $\displaystyle 12$ as 12x^0 $\displaystyle 12x^0$.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})$

Notice that $(0*12x^{0-1})=0$ $\displaystyle (0*12x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(10x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

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AP Calculus AB : Derivative as a function

Study concepts, example questions & explanations for AP Calculus AB

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Example Question #71 : Derivative As A Function

Example Question #72 : Derivative As A Function

Example Question #73 : Derivative As A Function

Example Question #11 : Equations Involving Derivatives

Example Question #681 : Derivatives

Example Question #1 : Finding Regions Of Concavity And Convexity

Example Question #1 : Finding Regions Of Concavity And Convexity

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