Derivatives

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AP Calculus BC › Derivatives

Questions 1 - 10

$g(x) = e ^{x^{2}+x}$

Evaluate .

$7 e^{12}$

$12 e^{12}$

$12 e^{11}$

$7 e^{9}$

$e^{7}$

Explanation

To find , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

$\begin{align*}&\text{The expression of a particular function is unknown;}\&\text{however, we have an expression for its derivative.}\&\text{Knowing that }f'(x)=4e^{(2x)}\text{ and }f(0)=9\&\text{Approximate } f(4.8)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Explanation

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\&\text{initial value, and step size are known, is:}\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\&\text{To calculate the step size find the distance between the final}\&\text{and initial value and divide by the number of steps to be used:}\&\Delta x = \frac{x_f-x_i}{Steps}\&\text{For this problem, we are told: }f'(x)=4e^{(2x)};f(0)=9\&\text{Knowing this, we may take the steps to estimate our function}\&\text{value at our final x value:}\&\Delta x = \frac{4.8-(0)}{4}=1.2\&y_0=9;x_0=0\&y_{1}=9+(1.2)(4e^{(2(0))})=13.8\&y_{2}=13.8+(1.2)(4e^{(2(1.2))})=66.71\&y_{3}=66.71+(1.2)(4e^{(2(2.4))})=649.96\&y_{4}=649.96+(1.2)(4e^{(2(3.6))})=7079.23\&\&\end{align*}$

Evaluate the limit using L'Hopital's Rule.

$\lim_{x\rightarrow \infty }\frac{x^2}{e^x+1}$

Undefined

$\infty$

Explanation

L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

Explanation

$g(x) = e ^{x^{2}+x}$

Evaluate .

$7 e^{12}$

$12 e^{12}$

$12 e^{11}$

$7 e^{9}$

$e^{7}$

Explanation

To find , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

Explanation

Evaluate the limit using L'Hopital's Rule.

$\lim_{x\rightarrow \infty }\frac{x^2}{e^x+1}$

Undefined

$\infty$

Explanation

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

$g(x) = e ^{x^{2}+x}$

Evaluate .

$7 e^{12}$

$12 e^{12}$

$12 e^{11}$

$7 e^{9}$

$e^{7}$

Explanation

To find , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

Evaluate the limit using L'Hopital's Rule.

$\lim_{x\rightarrow \infty }\frac{x^2}{e^x+1}$

Undefined

$\infty$

Explanation

$\lim_{x\rightarrow \infty }\frac{2x}{e^x}$ .

This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get

$\lim_{x\rightarrow \infty }\frac{2}{e^x}$ .

Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get

$\frac{2}{e^\infty }$ and $e^\infty=\infty$ .

So we can simplify the function by remembering that any number divided by infinity gives you zero.

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.

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\frac{3\pi}{4}$

$-\frac{\pi}{4}$

Explanation

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.

Meanvaluetheorem

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.

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