Applications of Derivatives - AP Calculus BC

Card 0 of 490

Evaluate:

Tap to see back →

and

Therefore, by L'Hospital's Rule, we can find by taking the derivatives of the expressions in both the numerator and the denominator:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}x \sin x = $\frac{\mathrm{d}$ }{\mathrm{d} x} x \cdot \sin x + $\frac{\mathrm{d}$ }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $2^{x}$ - 1 \right ) = $\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $e^{x \ln 2}$ - 1 \right )$

$= $\frac{\mathrm{d}$ }{\mathrm{d} x} $e^{x \ln 2}$ - $\frac{\mathrm{d}$ }{\mathrm{d} x} 1$

$= \ln 2\cdot $e^{x \ln 2}$ - 0 = \ln 2\cdot $2^{x }$$

$= $\frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot $2^{0 }$$} = $\frac{ 0 + 0 }{ \ln 2\cdot 1}$ = 0$

Evaluate:

Tap to see back →

Let's examine the limit

first.

$$\frac{\mathrm{d}$ }{\mathrm{d} x}x = 1$

and

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $2^{x}$ - 1 \right ) = $\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $e^{ x \ln 2}$ - 1 \right )$

$= $\frac{\mathrm{d}$ }{\mathrm{d} $x}e^{ x \ln 2}$ - $\frac{\mathrm{d}$ }{\mathrm{d} x} 1$

$=\ln 2 \cdot $e^{ x \ln 2}$ - 0 = \ln 2 \cdot $2^{x}$$ ,

so by L'Hospital's Rule,

Since ,

Now, for each , $0 \leq $\sin^{2}$ x \leq 1$ ; therefore,

$0 \leq $\frac{x $\sin^{2}$$ $x}{2^{x}$ - 1} \leq $\frac{x $}{2^{x}$$ - 1}$

By the Squeeze Theorem,

$0 \leq $\lim_{x\rightarrow \infty }$ $\frac{x $\sin^{2}$$ $x}{2^{x}$ - 1} \leq $\lim_{x\rightarrow \infty }$ $\frac{x $}{2^{x}$$ - 1} = 0$

and

Evaluate:

Tap to see back →

Therefore, by L'Hospital's Rule, we can find by taking the derivatives of the expressions in both the numerator and the denominator:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $5^{x}$ - 1 \right ) = $\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $e^{x \ln 5}$ - 1 \right )$

$= $\frac{\mathrm{d}$ }{\mathrm{d} x} $e^{x \ln 5 }$ - $\frac{\mathrm{d}$ }{\mathrm{d} x}1$

$=\ln 5\cdot $e^{x \ln 5 }$ - 0 = \ln 5\cdot $5^{x }$$

Similarly,

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $3^{x}$ - 1 \right ) = \ln 3 \cdot 3 ^{x}$

So

$=\frac{\ln 5 }{\ln 3 }$ \cdot $\lim_{x\rightarrow \infty }$ $\frac{ $5^{x}$$ }{ $3^{x}$ }= $\frac{\ln 5 }{\ln 3 }$ \cdot $\lim_{x\rightarrow \infty }$\left ( $\frac{ 5 }{ 3 }$ \right $)^{x}$$

But for any , so

Evaluate:

Tap to see back →

and

Therefore, by L'Hospital's Rule, we can find by taking the derivatives of the expressions in both the numerator and the denominator:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $7^{x}$ - 1 \right ) = $\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $e^{x \ln 7}$ - 1 \right )$

$= $\frac{\mathrm{d}$ }{\mathrm{d} x} $e^{x \ln 7 }$ - $\frac{\mathrm{d}$ }{\mathrm{d} x}1$

$=\ln 7\cdot $e^{x \ln 7 }$ - 0 = \ln 7\cdot $7^{x }$$

Similarly,

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\left ( $2^{x}$ - 1 \right ) = \ln 2 \cdot 2 ^{x}$

so

$= $\frac{ \ln 7 \cdot $7^{0 }$$}{ \ln 2\cdot $2^{0 }$} = $\frac{ \ln 7\cdot 1}{ \ln 2\cdot 1}$ = $\frac{ \ln 7}{ \ln 2 }$= $\log_{2}$7$

Suppose we have the following differential equation with the initial condition:

$$\frac{\partial p }{\partial x}$ = 0.5x(1-x), \ p(0) = 2$

Use Euler's method to approximate , using a step size of .

Tap to see back →

We start at x = 0 and move to x=2, with a step size of 1. Essentially, we approximate the next step by using the formula:

$p(x + \Delta x) = p(x) + p'(x)\cdot (\Delta x)$ .

So applying Euler's method, we evaluate using derivative:

$p'(x) = 0.5x(1-x), \ p(0) = 2$

And two step sizes, at x = 1 and x = 2.

${}p(x=1) = p(0) + (1 - 0)p'(0) = 2 + 1\cdot p'(0) = 2 + 1 \cdot 0.5(0)(1-0) = 2 + 0 = 2$

${}p(x=2) = p(1) + (2-1)\cdot p'(1) = 2 + 1\cdot p'(1) = 2 + 0.5(1)(1-1) = 2 + 0 = 2$

And thus the evaluation of p at x = 2, using Euler's method, gives us p(2) = 2.

Calculate the following limit.

Tap to see back →

To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get $\frac00$ , which is undefined.

What we can do to fix this is use L'Hopital's rule, which says

.

So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.

.

Plug in to get an answer of .

Approximate $\small $e^{$\frac{1}${2}$}$ by using Euler's method on the differential equation

$\small y'=y$

with initial condition $\small y(0)=1$ (which has the solution $\small $y(t)=e^t$$ ) and time step $\small \small h=\frac{1}{4}$$ .

Tap to see back →

Using Euler's method with $\small h=1/4$ means that we use two iterations to get the approximation. The general iterative formula is

$\small $y_{i+1}$=y_i+h\cdot f(t_i,y_i)$

where each $\small t_i$ is

$\small t_0=0$

$\small t_1=\frac{1}{4}$$

$\small t_2=\frac{1}{2}$$

$\small y_i$ is an approximation of $\small y(t_i)$ , and , for this differential equation. So we have

$\small \small \small \small y_1=y_0+\frac{1}{4}$f(t_0,y_0)=1+\frac{1}{4}$\cdot 1=1.25$

$\small \small \small \small $e^{1/2}$= y($\frac{1}{2}$)\approx y_2=y_1+\frac{1}{4}$f(t_1,y_1)=1.25+\frac{1}{4}$\cdot 1.25=1.5625$

So our approximation of $\small $e^{1/2}$$ is

$\small $e^{1/2}$\approx 1.5625$

Find the

$\small \small \small $\lim_{x\rightarrow $0}$$\frac{e^{x^{2}$$}-ln(1+x)-1}{cosx-sinx-1}$ .

Tap to see back →

Subbing in zero into $\small $$\frac{(e^{x^{2}$$}-ln(1+x)-1)}{cosx-sinx-1}$ will give you $\small $\frac{0}{0}$$ , so we can try to use L'hopital's Rule to solve.

First, let's find the derivative of the numerator.

$\small $e^{x^{2}$}$ is in the form $\small $a^{u}$$ , which has the derivative $\small $a^{u}$u'lna$ , so its derivative is $\small $2xe^{x^{2}$}$ .

$\small ln(1+x)$ is in the form $\small lnu$ , which has the derivative $\small u'/u$ , so its derivative is $\small 1/(1+x)$ .

The derivative of $\small 1$ is $\small 0$ so the derivative of the numerator is $\small $2xe^{x^{2}$}-1/(1+x)$ .

In the denominator, the derivative of $\small cosx$ is $\small -sinx$ , and the derivative of $\small sinx$ is $\small cosx$ . Thus, the entire denominator's derivative is $\small -(sinx+cosx)$ .

Now we take the

$\small $\lim_{x\rightarrow $0}$$\frac{2xe^{x^{2}$$}-1/(x+1)}{-(sinx+cosx)}$ , which gives us $\small $\frac{-1}{-1}$=1$ .

Evaluate the limit using L'Hopital's Rule.

Tap to see back →

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

.

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

.

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.

Evaluate the limit using L'Hopital's Rule.

Tap to see back →

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

.

Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.

This gives us

$4\cdot $0^2$=0$ .

Evaluate the limit using L'Hopital's Rule.

Tap to see back →

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

.

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

.

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

$5\cdot \infty=\infty$ .

Calculate the following limit.

$\lim_${x->-\infty}$\frac{e^{-2x}$$$}{x^2$}$

Tap to see back →

If we plugged in $-\infty$ directly, we would get an indeterminate value of $\frac{\infty}{\infty}$ .

We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.

$\lim_${x->-\infty}$\frac{e^{-2x}$$$}{x^2$$}=-$\frac{e^{-2x}$$}{x}$ .

We still can't evaluate the limit of the new expression, so we do it one more time.

$\lim_${x->-\infty}-$\frac{e^{-2x}$$}{x}=\lim_${x->-\infty}2e^{-2x}$=\infty$

Evaluate the following limit:

Tap to see back →

When you try to solve the limit using normal methods, you find that the limit approaches zero in the numerator and denominator, resulting in an indeterminate form "0/0".

In order to evaluate the limit, we must use L'Hopital's Rule, which states that:

when an indeterminate form occurs when evaluting the limit.

Next, simply find f'(x) and g'(x) for this limit:

$2t, -\cos(2-t)$

The derivatives were found using the following rules:

, $\frac{d}{dx}$\sin(u)=\cos(u)$\frac{du}{dx}$

Next, using L'Hopital's Rule, evaluate the limit using f'(x) and g'(x):

Find the limit if it exists.

Hint: Apply L'Hospital's Rule.

Tap to see back →

Through direct substitution, we see that the limit becomes

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit

is in indeterminate form, then the limit is equivalent to

Taking the derivatives we use the power rule which states

$\frac{\mathrm{d}$ }{\mathrm{d} $x}x^n$$=nx^{n-1}$

Using the power rule the limit becomes

$= $\lim_{x\rightarrow 0}$$\frac{4x-3}{6x+3}$$

As such the limit exists and is

Find the limit if it exists.

Hint: Apply L'Hospital's Rule.

Tap to see back →

Through direct substitution, we see that the limit becomes

which is in indeterminate form.

As such we can use l'Hospital's Rule, which states that if the limit

is in indeterminate form, then the limit is equivalent to

Taking the derivatives we use the trigonometric rule which states

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(Bx)=B\cos(Bx)$

where is a constant.

Using l'Hospital's Rule we obtain

And through direct substitution we find

$=\frac{\pi \cdot1}{1}$$

$=\pi$

As such the limit exists and is

Find the limit:

Tap to see back →

By substituting the value of , we will find that this will give us the indeterminate form $\frac{0}{0}$ . This means that we can use L'Hopital's rule to solve this problem.

L'Hopital states that we can take the limit of the fraction of the derivative of the numerator over the derivative of the denominator. L'Hopital's rule can be repeated as long as we have an indeterminate form after every substitution.

Take the derivative of the numerator.

Take the derivative of the numerator.

$$\frac{d}{dx}$(x-3) =1$

Rewrite the limit and use substitution.

The limit is .

Find the limit if it exists

Hint: Use L'Hospital's rule

Tap to see back →

Directly evaluating for $x=\infty$ yields the indeterminate form

$\frac{\infty ^2+3(\infty )}{2+ \infty}$=\frac{\infty}{\infty}$

we are able to apply L'Hospital's rule which states that if the limit is in indeterminate form when evaluated, then

As such the limit in the problem becomes

Evaluating for $x=\infty$ yields

$1+\frac{3}{2(\infty)}$ = 1+0 =1$

As such

and thus

Evaluate using L'hopital's rule.

Tap to see back →

This important limit from elementary limit theory is usually proven using trigonometric arguments, but it can be shown using L'hopital's rule too.

Evaluate the limit:

Tap to see back →

When evaluating the limit using normal methods, we find that the indeterminate form $\frac{0}{0}$ results. When this occurs, we must use L'Hopital's Rule, which states that for .

Taking the derivative of the top and bottom functions and evaluating the limit, we get

The derivatives were found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} ax=a$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(x)=\cos(x)$

Evaluate the limit:

Tap to see back →

When evaluating the limit using normal methods (substitution), we get the indeterminate form $\frac{0}{0}$ . When this (or $\frac{\infty}{\infty}$ ) occurs, we must use L'Hopital's Rule to evaluate the limit. The rule states that when an indeterminate form results from the limit,

Taking the derivatives in our limit, we get

The derivatives were found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} ax=a$