Derivatives - AP Calculus BC

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Question

Let on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval

Then take the difference of the two and divide by the interval.

$\frac{43-3}{2-0}=20$

Now find the derivative of the function; this will be solved for the value(s) found above.

Since the interval is , satisfies the MVT.

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Question

Let on the interval $(0,\frac{\pi}{2})$ . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval $(0,\frac{\pi}{2})$

$f(\frac{\pi}{2})=cos(4(\frac{\pi}{2}))=cos(2\pi)=1$

Then take the difference of the two and divide by the interval.

$\frac{1-1}{\frac{\pi}{2}-0}=0$

Now find the derivative of the function; this will be solved for the value(s) found above.

$x=\frac{sin^{-1}(0)}{4}$

Multiple solutions will solve this function, but on the interval $(0,\frac{\pi}{2})$ , only $x=\frac{\pi}{4}$ fits within, satisfying the MVT.

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Question

Let on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval

Then take the difference of the two and divide by the interval.

$\frac{40-(-5)}{3-0}=\frac{45}{3}=15$

Now find the derivative of the function; this will be solved for the value(s) found above.

$x=\frac{3}{2}$ , which falls within the interval , satisfying the MVT.

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Question

Let on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval

Then take the difference of the two and divide by the interval.

$\frac{55-9}{5-3}=\frac{46}{2}=23$

Now find the derivative of the function; this will be solved for the value(s) found above.

which falls within the interval , satisfying the MVT.

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Question

Let on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval

Then take the difference of the two and divide by the interval.

$\frac{238-150}{0.1-0}=880$

Now find the derivative of the function; this will be solved for the value(s) found above.

, which falls between , satisfying the MVT.

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Question

Let on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of on the interval

Then take the difference of the two and divide by the interval.

$\frac{ln(7)-ln(6)}{4-3}=0.154$

Now find the derivative of the function; this will be solved for the value(s) found above.

$d(ln(u))=\frac{du}{u}$

$f'(x)=\frac{1}{x+3}$

$\frac{1}{x+3}=0.154$

$x+3=\frac{1}{0.154}$

which falls within the interval satisfying the mean value theorem.

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Question

Let $f(x)=e^{-\frac{1}{x}}$ on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of $f(x)=e^{-\frac{1}{x}}$ on the interval

$f(1)=e^{-\frac{1}{1}}=e^{-1}$

$f(3)=e^{-\frac{1}{3}}$

Then take the difference of the two and divide by the interval.

$\frac{e^{-\frac{1}{3}}-e^{-1}}{3-1}=0.174$

Now find the derivative of the function; this will be solved for the value(s) found above.

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

$\frac{1}{x^2}e^{-\frac{1}{x}}=0.174$

There are multiple solutions; within the interval , satisfies the mean value theorem.

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Question

Let $f(x)=16^{\frac{1}{x}}$ on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of $f(x)=16^{\frac{1}{x}}$ on the interval

$f(2)=16^{\frac{1}{2}}=4$

$f(4)=16^{\frac{1}{4}}=2$

Then take the difference of the two and divide by the interval.

$\frac{2-4}{4-2}=-1$

Now find the derivative of the function; this will be solved for the value(s) found above.

$f'(x)=-\frac{1}{x^2}16^{\frac{1}{x}}ln(16)$

$-\frac{1}{x^2}16^{\frac{1}{x}}ln(16)=-1$

This solution falls within , validating the mean value theorem.

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Question

Let $f(x)=xe^{4x}$ on the interval . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Answer

The mean value theorem states that for a planar arc passing through a starting and endpoint , there exists at a minimum one point, , within the interval for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of $f(x)=xe^{4x}$ on the interval

$f(0)=(0)e^{4(0)}=0$

$f(1)=(1)e^{4(1)}=54.6$

Then take the difference of the two and divide by the interval.

$\frac{54.6-0}{1-0}=54.6$

Now find the derivative of the function; this will be solved for the value(s) found above.

Derivative of an exponential:

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Product rule:

$f'(x)=e^{4x}+4xe^{4x}$

$e^{4x}+4xe^{4x}=54.6$

Using a calculator, we find the solution , which fits within the interval , satisfying the mean value theorem.

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Question

Find $\frac{dy}{dx}$ if $x^3 -x\ln y=ye^x$ .

Answer

This function is implicit, because y is not defined directly in terms of only x. We could try to solve for y, but that would be difficult, if not impossible. The easier solution would be to employ implicit differentiation. Our strategy will be to differentiate the left and right sides by x, apply the rules of differentiation (such as Chain and Product Rules), group dy/dx terms, and solve for dy/dx in terms of both x and y.

$x^3-x\ln y=ye^{x}$

$\frac{d}{dx}[x^3-x\ln y]=\frac{d}{dx}[ye^x]$

$\frac{d}{dx}[x^3]-\frac{d}{dx}[x\ln y]=\frac{d}{dx}[ye^x]$

We will need to apply both the Product Rule and Chain Rule to both the xlny and the terms.

According to the Product Rule, if f(x) and g(x) are functions, then $\frac{d}{dx}[f(x)g(x)]=f(x)\cdot \frac{d}{dx}[g(x)]+g(x)\cdot \frac{d}{dx}[f(x)]$ .

And according to the Chain Rule,

$\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)$

$3x^2-(x\cdot \frac{d}{dx}[\ln y]+\ln y\cdot \frac{d}{dx}[x])=e^x\cdot \frac{d}{dx}[y]+y\cdot \frac{d}{dx}[e^x]$

$3x^2-(x\cdot \frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{d}{dy}[y]\cdot \frac{dy}{dx}+y\cdot e^x$

$3x^2-(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{dy}{dx}+ye^x$

$3x^{2}-\frac{x}y{}\cdot \frac{dy}{dx}-\ln y=e^x\cdot \frac{dy}{dx}+ye^x$

Now we will group the dy/dx terms and move everything else to the opposite side.

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

Then, we can solve for dy/dx.

$\frac{3x^2-\ln y-ye^x}{e^x+\frac{x}{y}} =\frac{dy}{dx}$

To remove the compound fraction, we can multiply the top and bottom of the fraction by y.

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

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

The (ugly) answer is $\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$ .

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Question

$f (x) = 7x^{2} + 6x - 9$

Give .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

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Question

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Question

Differentiate $x^{999}$ .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Question

Give the second derivative of $x^{777}$ .

Answer

Find the derivative of $x^{777}$ , then find the derivative of that expression.

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

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Question

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Answer

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

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Question

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give .

Answer

First, find the derivative of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate to get .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Question

Find the derivative of:

Answer

The derivative of inverse cosine is:

$\frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

$\frac{d}{dx} cos(x)=-sin(x)$

Combine the two terms into one term.

$-\frac{1}{\sqrt{1-x^2}}-sin(x)$

$-\frac{1}{\sqrt{1-x^2}}-\frac{sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}= -\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

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Question

What is the rate of change of the function at the point ?

Answer

The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is

$\frac{d}{dx}x^n=nx^{n-1}$ , and that the derivative of a constant is zero.

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

Therefore, the rate of change of f(x) at the point (1,6) is 14.

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Question

Find the derivative of the function $y = \int_{0}^{x^2} e^{-t^2}dt$

Answer

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$y = \int_{0}^{x^2} e^{-t^2}dt$ . Start

$y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$ . Take the derivative of both sides with respect to .

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's in this case), and that the upper limit of the integral is a function of , (it's in this case).

Afterward, plug in for , and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$ .

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Question

Find the derivative of the function $y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$

Answer

To find the derivative of this function, we need to use the Fundemental Theorem of Calculus Part 1 (As opposed to the 2nd part, which is what's usually used to evaluate definite integrals)

$y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Start

$y'= \frac{d}{dx}\int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Take derivatives of both sides.

$y'= x^2e^x\ln(x)\sin(x)$ . "Cancel" the integral and the derivative. (Make sure that the upper bound on the integral is a function of , and that the lower bound is a constant before you cancel, otherwise you may need to use some manipulation of the bounds to make it so.)

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