Derivatives - AP Calculus BC

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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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Question

Given $f(x)=5x^{2}+6x-12$ , find the value of at the point .

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Answer

Given the function $f(x)=5x^{2}+6x-12$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6$ .

Plugging in the -value of the point into , we get

.

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Question

Given $f(x)=10x^{2}-9x+1$ , find the value of at the point .

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Answer

Given the function $f(x)=10x^{2}-9x+1$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)10x^{2-1}-(1)9x^{1-1}+(0)=20x-9$ .

Plugging in the -value of the point into , we get

.

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Question

Find the derivative of at point .

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Answer

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

Substitute .

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Question

Find the derivative of the following function at :

$h(x)=2x\sin(x)$

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Answer

The derivative of the function is given by the product rule:

,

Simply find the derivative of each function:

$\frac{d}{dx}(2x)=2$

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

The derivatives were found using the following rules:

$\frac{d}{dx}\sin(x)=\cos(x)$ , $\frac{d}{dx}(x^n)=nx^{n-1}$

Simply evaluate each derivative and the original functions at the point given, using the above product rule.

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Question

What is the slope of a function at the point ?

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Answer

Slope is defined as the first derivative of a given function.

Since , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

$f'(x)=(2)(-2)x^{2-1}+(1)5x^{1-1}-(0)3=-4x+5$ .

Since we're given a point , we can use the x-coordinate to solve for the slope at that point.

Thus,

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Question

What is the slope of the tangent line to the function

$h(x)=e^{x}ln(x)$

when

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Answer

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

$\frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)$ .

In this example, $m(x)=e^x \ and\ n(x)=ln(x).$

Therefore,

$m'(x)=e^x \ and\ n'(x)=\frac{1}{x}$ , and

$\frac{d}{dx}(e^xln(x))=e^xln(x)+e^x\cdot\frac{1}{x}$

At x = 1, this dervative has the value

$e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e$ .

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Question

Find the second derivative of the following function at :

$e^{2x}+2x\cos(3x)$

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Answer

To find the second derivative of the function, we first must find the first derivative of the function:

$f'(x)=2e^{2x}+2\cos(3x)-6x\sin(3x)$

The derivative was found using the following rules:

$\frac{d}{dx}\cos(x)=-\sin(x)$ , $\frac{d}{dx}e^u=e^u\frac{du}{dx}$ , $\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

The second derivative is simply the derivative of the first derivative function, and is equal to:

$f''(x)=4e^{2x}-12\sin(3x)-18x\cos(3x)$

One more rule used in combination with some of the ones above is:

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

To finish the problem, plug in x=0 into the above function to get an answer of .

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Question

Find for

$f(x)=x^3\ln(x+1)$

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Answer

In order to find the derivative, we need to find . We can find this by remembering the product rule and knowing the derivative of natural log.

Product Rule:

.

Derivative of natural log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Now lets apply this to our problem.

$f(x)=x^3\ln(x+1)$

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

$f'(0)=3(0)^2\ln(0+1)+\frac{0^3}{0+1}=0$

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Question

Calculate the derivative of $f(x)=\frac{x^3}{3}-2x^2+x-3$ at the point .

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Answer

There are 2 steps to solving this problem.

First, take the derivative of

Then, replace the value of x with the given point and evaluate

For example, if , then we are looking for the value of , or the derivative of at .

$f(x)=\frac{x^3}{3}-2x^2+x-3$

Calculate

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$

Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

Then, plug in the value of x and evaluate

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Question

If $f(x)=\ln(2x)$ , which of the following limits equals ?

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Answer

The equation for the derivative at a point is given by

$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ .

By substituting , $f = \ln(2x)$ , we obtain

$f'(e+1)=\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}.$

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