Derivatives - AP Calculus BC

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Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

← Didn't Know|Knew It →

Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Tap to reveal answer

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

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

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

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

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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Question

The position of an object is modeled by the equation $y= \frac{1}{2} \cos ^2 t$ What is the speed after $\frac{\pi}{4}$ seconds?

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Answer

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: $f(g(x)) ' = f'(g(x))\cdot g'(x)$ . In this case, $f(x) = \frac{1}{2} x^2$ and $g(x) = \cos x$ . Since and $g'(x ) = - \sin (x)$ , the first derivative is $\cos t \cdot - \sin t$ .

Plug in $\frac{\pi}{4}$ for t:

$\cos (\frac{\pi}{4} ) \cdot - \sin (\frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt2} = -\frac{1}{2}$

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Question

$g(x) = \sec \frac{\pi}{x}$

Evaluate .

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Answer

To find , substitute $u =\frac{ \pi }{x} = \pi x ^{-1}$ and use the chain rule:

$g(x) = \sec \left (\frac{\pi}{x} \right )$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \sec\frac{\pi}{x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} \sec u$

$= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d}u} \sec u$

$= \frac{\mathrm{d} }{\mathrm{d} x} \pi x^{-1} \cdot \frac{\mathrm{d} } {\mathrm{d}u} \sec u$

$= \pi \cdot (-1) \cdot x^{-1-1} \cdot \tan u \sec u$

$= - \pi x^{-2} \cdot \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

$= - \frac{\pi}{x^{2}} \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

So $g '(x)= - \frac{\pi}{x^{2}}\tan \frac{\pi}{x} \sec \frac{\pi}{x}$

and

$g '(6)= - \frac{\pi}{6^{2}}\tan \frac{\pi}{6} \sec \frac{\pi}{6}$

$= - \frac{\pi}{36} \cdot \frac{\sqrt{3}}{3} \cdot \frac{2\sqrt{3}}{3}$

$= - \frac{\pi}{36} \cdot \frac{2\cdot 3}{9} = - \frac{\pi}{54}$

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Question

What is the equation of the line tangent to the graph of the function

$f(x)= 1+\frac{4}{x}$

at the point ?

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Answer

The slope of the line tangent to the graph of $f(x)= 1+\frac{4}{x}$ at is

, which can be evaluated as follows:

$f(x)= 1+\frac{4}{x} = 1 + 4x^{-1}$

$f'(x) = (-1) 4x^{-1-1}$

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

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

The equation of the line with slope through is:

$y - y_{1} = m (x-x_{1})$

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Question

What is the equation of the line tangent to the graph of the function

$f(x) = 2 \sin \frac{\pi x}{6}$

at the point ?

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Answer

The slope of the line tangent to the graph of $f(x) = 2 \sin \frac{\pi x}{6}$ at is

, which can be evaluated as follows:

$f(x) = 2 \sin \frac{\pi x}{6} = 2 \sin \frac{\pi }{6}x$

$f'(x) = 2 \cdot\frac {\pi }{6} \cos \frac{\pi }{6}x$

$f'(x) = \frac {\pi }{3} \cos \frac{\pi }{6}x$

$f'(7) = \frac {\pi }{3} \cos \frac{7 \pi }{6}$

$= \frac {\pi }{3} \left ( - \frac{\sqrt{3}}{2}\right ) = - \frac {\pi \sqrt{3 }}{6}$ , the slope of the line.

The equation of the line with slope $- \frac {\pi \sqrt{3 }}{6}$ through is:

$y - y_{1} = m (x-x_{1})$

$y - (-1) = - \frac {\pi \sqrt{3 }}{6} (x-7)$

$y +1 = - \frac {\pi \sqrt{3 }}{6} x+\left ( \frac {\pi \sqrt{3 }}{6} \right ) 7$

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

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Question

What is the equation of the line tangent to the graph of the function

$f(x)= \frac{8}{x}$

at ?

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Answer

The slope of the line tangent to the graph of $f(x)= \frac{8}{x}$ at is

, which can be evaluated as follows:

$f(x)= \frac{8}{x} = 8x^{-1}$

$f'(x)= 8 \cdot\left ( -1 x^{-2} \right )$

$f'(x)= -\frac{8}{x^{ 2}}$

$f'(-2)= -\frac{8}{(-2)^{ 2}} = -\frac{8}{4} = -2$ , the slope of the line.

The equation of the line with slope through is:

$y - y_{1} = m (x-x_{1})$

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Question

What is the equation of the line tangent to the graph of the function

$f(x)= \log (x+3)$

at the point ?

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Answer

The slope of the line tangent to the graph of $f(x)= \log (x+3)$ at the point is , which can be evaluated as follows:

$f(x)= \log (x+3)$

$f(x)=\frac{ \ln (x+3) }{\ln 10}$

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \frac{ \ln (x+3) }{\ln 10}$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{\mathrm{d} }{\mathrm{d} x}\ln (x+3)$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{1}{x+3}$

$f'(7)= \frac{1 }{\ln 10} \cdot \frac{1}{7+3}$

$f'(7)= \frac{1 }{10 \ln 10}$

The line with this slope through has equation:

$y - y_{1} = m (x-x_{1})$

$y - 1= \frac{1 }{10 \ln 10} (x-7)$

$y - 1= \frac{1 }{10 \ln 10} \cdot x- \frac{1 }{10 \ln 10} \cdot 7$ $y = \frac{x }{10 \ln 10} - \frac{7 }{10 \ln 10} +1$

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

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Question

What is the equation of the line tangent to the graph of the function

$f(x) = x^{3} + x - 17$

at the point ?

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Answer

The slope of the line tangent to the graph of $f(x) = x^{3} + x - 17$ at the point is , which can be evaluated as follows:

$f(x) = x^{3} + x - 17$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )$

$f'(x) = 3x^{2} + 1$

$f'(3) = 3\cdot 3^{2} + 1 = 28$

The line with slope 28 through has equation:

$y - y_{1} = m (x-x_{1})$

$y -13= 28 x- 28 \cdot 3$

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Question

Given the function , find the slope of the point .

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Answer

To find the slope at a point of a function, take the derivative of the function.

The derivative of is $\frac{1}{x:ln(a)}$ .

Therefore the derivative becomes,

$\frac{1}{x:ln(e)}+5=\frac{1}{x}+5$ since .

Now we substitute the given point to find the slope at that point.

$\frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}$

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Question

Find the value of the following derivative at the point $\left(2,\frac{1}{2}\right)$ :

$f(x)=\frac{1}{\sqrt{x+2}}$

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Answer

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

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

From here we need to evaluate at the given point $\left(2,\frac{1}{2}\right)$ . In this case, only the x value is important, so we evaluate our derivative at x=2 to get $f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$ .

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Question

Evaluate the value of the derivative of the given function at the point :

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Answer

To solve this problem, first we need to take the derivative of the function.

$\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\f'(x)=12x^3+4x$

From here we need to evaluate at the given point . In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\f'(-1)=12(-1)^3+4(-1)\ f'(-1)=12(-1)+4(-1)\ f'(-1)=-16$ .

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