Understanding the limiting process. - AP Calculus AB

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Question

Find the slope of the tangent line to the graph of f at x = 9, given that f(x) = –x2 + 5√(x)

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Answer

First find the derivative of the function.

f(x) = –x2 + 5√(x)

f'(x) = –2x + 5(1/2)x–1/2

Simplify the problem

f'(x) = –2x + (5/2x1/2)

Plug in 9.

f'(3) = –2(9) + (5/2(9)1/2)

= –18 + 5/(6)

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Question

Find the derivative

(x + 1)/(x – 1)

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Answer

Rewrite problem.

(x + 1)/(x – 1)

Use quotient rule to solve this derivative.

((x – 1)(1) – (x + 1)(1))/(x – 1)2

(x – 1) – x – 1)/(x – 1)2

–2/(x – 1)2

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Question

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

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Answer

Use the chain rule and the formula

$\frac{d}{dx}sin^{-1}u = \frac{1}{\sqrt{1-x^2}}$

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Question

Find the derivative.

y = sec (5x3)

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Answer

The derivative of the function y = sec(x) is sec(x)tan(x). First take the derivative of the outside of the function: y = sec(4x3) : y' = sec(5x3)tan(5x3). Then take the derivative of the inside of the function: 5x3 becomes 15x2. So your final answer is: y' = ec(5x3)tan(5x3)15x2

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Question

What is the derivative of $(2+3cos(3x))^\pi$ ?

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Answer

Need to use the power rule which states: $\frac{d}{dx}u^n=nu^{n-1}\frac{du}{dx}$

In our problem $\frac{du}{dx}=-3sin(3x)$

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Question

Find the derivative of

$f(x)=4(x^{3}-2x+11)(x-2)$

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Answer

The answer is $f'(x)=16x^{3}-24x^{2}-16x+60$ . It is easy to solve if we multiply everything together first before taking the derivative.

$f(x)=4(x^{3}-2x+11)(x-2)$

$f(x)=4(x^{4}-2x^{3}+-2x^{2}+4x+11x-22)$

$f(x)=4x^{4}-8x^{3}-8x^{2}+60x-22$

$f'(x)=16x^{3}-24x^{2}-16x+60$

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Question

If $y=x^{3}e^{x}$ , then $\frac{dy}{dx} = ?$

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Answer

The correct answer is $x^{2}e^{x}(3+x)$ .

We must use the product rule to solve. Remember that the derivative of $e^{x}$ is $e^{x}$ .

$y'=(3x^{2})(e^{x})+(x^{3})(e^{x})$

$y'=x^{2}e^{x}(3+x)$

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Question

$\lim_{x\rightarrow 2^-}=\frac{x^2-4}{x-2}$

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Answer

This question is asking to evaluate a one-sided equation of a function. Specifically, the limit of the function to the left of . When is substituted into the function the result is indeterminate. This means it is in the form zero over zero.

Since the question is looking for a one-sided limit, let us substitute in a value that is slightly less than .

$\lim_{x\rightarrow 2^-}=\frac{x^2-4}{x-2}$

$\\lim_{x=1.9}\frac{1.9^2-4}{1.9-2} \\=\frac{3.61-4}{1.9-2} \\=\frac{-0.39}{-0.1} \\=3.9$

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Question

Find $\frac{d^{2}y}{dx^{2}}$ if .

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Answer

We will have to find the first derivative of with respect to using implicit differentiation. Then, we can find $\frac{d^{2}y}{dx^{2}}$ , which is the second derivative of with respect to .

$\frac{d}{dx}[y^2-x^2]=\frac{d}{dx}[4]$

We will apply the chain rule on the left side.

$\frac{d}{dx}[y^2]\cdot \frac{dy}{dx}-\frac{d}{dx}[x^2]=0$

$2y\cdot \frac{dy}{dx}-2x=0$

We now solve for the first derivative with respect to .

$\frac{dy}{dx}=\frac{x}{y}$

In order to get the second derivative, we will differentiate $\frac{x}{y}$ with respect to . This will require us to employ the Quotient Rule.

$\frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[\frac{x}{y}]=\frac{y\cdot \frac{d}{dx}[x]-x\cdot \frac{d}{dx}[y]}{y^{2}} =\frac{y\cdot 1-x\cdot \frac{dy}{dx}}{y^{2}}$

We will replace $\frac{\mathrm{d}y }{\mathrm{d} x}$ with $\frac{x}{y}$ .

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

But, from the original equation, $y^{2}-x^{2}=4$ . Also, if we solve for , we obtain $y = (4+x^{2})^{\frac{1}{2}}$ .

$\frac{d^{2}y}{dx^{2}}=\frac{y^{2}-x^{2}}{y^{3}}=\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

The answer is $\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$ .

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Question

Differentiate $y=e^{x}+x^e$ .

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Answer

The derivative of is equal to therefore the first part of the equation remains the same.

The second part requires regular differential rules.

$x^ndx=nx^{n-1}$

Therefore when differentiating you get $ex^{e-1}$ .

Combining the first and second part we get the final derivative:

$y'=e^x+ex^{e-1}$ .

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Question

Differentiate $y=\sin^{-1}(4x)$ .

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Answer

The rule for taking the derivative of $sin^{-1}(x)=\frac{1}{\sqrt{1-x^2}}$ .

For this problem we need to remember to use the Chain Rule.

Since we are taking the derivative of,

$sin^{-1}(4x)$ we need to take the derivative of the outside piece $sin^{-1}$ keeping the inside piece the same , and then multiply the whole thing by the derivative of the inside piece .

Therefore the solution becomes:

$y'=\frac{1}{\sqrt{1-16x^2}}\cdot 4$ ,

$y'=\frac{4}{\sqrt{1-16x^2}}$ .

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Question

Evaluate:

$\int\frac{1}{x}dx$ .

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Answer

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

Therefore the integral of

$\int\frac{1}{x}dx= \ln(x)+C$ where C is some constant.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow 5}\frac{x^2-2x-63}{x-9}$

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Answer

Normally, you would factor out x-9 from the numerator and denominator, but it isn't necessary for this problem. Since the problem asks for the limit to be evaluated at x=5 where there is no discontinuity, the limit will be f(5).

$f(5)=\frac{(5)^2-2(5) -63}{5-9}=12$

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Question

If what is ?

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Answer

The derivative of,

$\ln(1x)=\frac{1}{x}\cdot 1$ .

The derivative of

$\ln(2x)= \frac{1}{2x}\cdot2$

Therefore, using the Chain Rule the derivative of the function will become the derivative of the outside piece keeping the original inside piece. Then multiplying that by the derivative of the inside piece.

$\ln(2x+6)= \frac{1}{2x+6}\cdot2$

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Question

Differentiate .

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Answer

Using the power rule, multiply the coefficient by the power and subtract the power by 1.

$f'(x)=5*3(x^{3-1})$

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Question

Differentiate $f(x)=\frac{1}{x^2}$ .

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Answer

Use the product rule:

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

$f'(x)=-21(x^{-2-1})= {-2}{x^{-3}}= \frac{-2}{x^3}$

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Question

Differentiate:

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Answer

Use the product rule to find the derivative of the function.

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Question

Differentiate: $y=e^{x^2}$

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Answer

The derivative of any function of e to any exponent is equal to the function multiplied by the derivative of the exponent.

$f'(x)=e^{x^2}*2x$

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Question

Find the second derivative of $y=ln(\frac{x^2}{x^2+1})$ .

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Answer

$y'= \frac{1}{x^2}2x-\frac{1}{x^2+1}2x$

$\frac{2x}{x^2}-\frac{2x}{x^2+1}= \frac{2x^3+2x-2x^3}{x^2(x^2+1)}= \frac{2x}{x^4+x^2}$

Factoring out an x gives you $\frac{2}{x^3+x}$ .

$y''=2(x^3+x)^{-1}=-2(x^3+x)^{-2}(3x^2+1)=\frac{-2(3x^2+1)}{(x^3+x)^2}$

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Question

Consider:

$f(x) = x^{99} - x^{66}$

The 99th derivative of is:

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Answer

For $f(x) = x^{n}$ , the nth derivative is . As an example, consider $f(x) = x^{3}$ . The first derivative is $3x^{2}$ , the second derivative is , and the third derivative is . For the question being asked, the 99th derivative of $x^{99}$ would be . The 66th derivative of $x^{66}$ would be , and any higher derivative would be zero, since the derivative of any constant is zero. Thus, for the given function, the 99th derivative is .

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