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Calculus AB : Determine Limits Using Algebraic Properties and the Squeeze Theorem

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Example Question #1 : Calculus Ab

Given:

Find $f\left ( x \right )$

Possible Answers:

(x^2-5x+2)^4(36x^3-45x^2+12x)

(x^2-5x+2)^4(36x^3-105x^2+12x)

(x^2-5x+2)^5(36x^3-105x^2+12x)

(15x^2)(x^2-5x+2)^4

(x^2-5x+2)^4(60x^2-150)

Correct answer:

(x^2-5x+2)^4(36x^3-105x^2+12x)

Explanation:

Computation of the derivative requires the use of the Product Rule and Chain Rule.

The Product Rule is used in a scenario when one has two differentiable functions multiplied by each other:

f(x)*g(x)

$\frac{d}{dx}[f(x)*g(x)]=f(x)*g'(x)+g(x)*f(x)$

This can be easily stated in words as: "First times the derivative of the second, plus the second times the derivative of the first."

In the problem statement, we are given:

f(x)=3x^2(x^2-5x+2)^5

3x^2 is the "First" function, and (x^2-5x+2)^5 is the "Second" function.

The "Second" function requires use of the Chain Rule.

When:

$y=(f(x))^a \rightarrow \frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these formulas results in:

f'(x)=(3x^2)[5(x^2-5x+2)^4(2x-5)]+(x^2-5x+2)^5(6x)

Simplifying the terms inside the brackets results in:

f'(x)=(3x^2)[(10x-25)(x^2-5x+2)^4]+(x^2-5x+2)^5(6x)

We notice that there is a common term that can be factored out in the sets of equations on either side of the "+" sign. Let's factor these out, and make the equation look "cleaner".

f'(x)=(x^2-5x+2)^4[(3x^2)(10x-25)+(6x)(x^2-5x+2)]

Inside the brackets, it is possible to clean up the terms into one expanded function. Let us do this:

f'(x)=(x^2-5x+2)^4[30x^3-75x^2+6x^3-30x^2+12x]

Simplifying this results in one of the answer choices:

f'(x)=(x^2-5x+2)^4(36x^3-105x^2+12x)

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Example Question #2 : Calculus Ab

$f(x)=x^2\tan^{-1}(3x^5)$

$\frac{dy}{dx}=?$

Possible Answers:

$\frac{15x^8}{1+9x^7}+2x\tan^{-1}(3x^5)$

$2x\tan^{-1}(3x^5)$

$\frac{15x^4\tan^{-1}(3x^5)}{1+9x^{7}}+2x^3$

$\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

$\frac{15x^4\tan^{-1}(3x^5)}{1+9x^{10}}+2x^3$

Correct answer:

$\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

Explanation:

Evaluation of this integral requires use of the Product Rule. One must also need to recall the form of the derivative of $\tan^{-1}(x)$ .

Product Rule:

$\frac{d}{dx}[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)$

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

Applying these two rules results in:

$\frac{dy}{dx}=\frac{d}{dx}[x^2\tan^{-1}(3x^5)]$

$\frac{dy}{dx}=(x^2)(\frac{1}{1+(3x^5)^2})(15x^4)+\tan^{-1}(3x^5)(2x)$

$\frac{dy}{dx}=\frac{15x^6}{1+9x^{10}}+2x\tan^{-1}(3x^5)$

This matches one of the answer choices.

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Example Question #3 : Calculus Ab

Complete the derivative:

$f(x)=2x^2\tan^{-1}(3x)+5\ln(10x)$

Possible Answers:

$f'(x)=\frac{6x^2}{1+3x^2}+4x\tan^{-1}(3x)+\frac{100}{x}$

$f'(x)=\frac{6x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

$f'(x)=\frac{3x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

$f'(x)=\frac{3x^2}{1+3x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

$f'(x)=\frac{2x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{50}{x}$

Correct answer:

$f'(x)=\frac{6x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

Explanation:

Computation of this derivative will require the use of the Product Rule, and knowledge of the derivative of the inverse tangent function, and natural logarithmic function:

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

$\frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}$

We can now easily compute the derivative.

$f'(x)=(2x^2)(\frac{1}{1+(3x)^2})(3)+(\tan^{-1}(3x))(4x)+5(\frac{1}{10x})(10)$

This simplifies to:

$f'(x)=\frac{6x^2}{1+9x^2}+4x\tan^{-1}(3x)+\frac{5}{x}$

This is one of the answer choices.

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Example Question #4 : Calculus Ab

Find dy/dx:

$y=\tan^{-1}(2x(x+1)^{10})$

Possible Answers:

$\frac{dy}{dx}=2(x+1)^9[\frac{11x+1}{1+4x^2(x+1)^{12}}]$

$\frac{dy}{dx}=\frac{11x+1}{1+4x^2(x+1)^{20}}$

$\frac{dy}{dx}=2(x+1)^{10}[\frac{11x+1}{1+4x^2(x+1)^{12}}]$

$\frac{dy}{dx}=2(x+1)^9[\frac{11x+1}{1+4x^2(x+1)^{20}}]$

$\frac{dy}{dx}=2(x+1)^{10}[\frac{10x+1}{1+4x^2(x+1)^{2}}]$

Correct answer:

$\frac{dy}{dx}=2(x+1)^9[\frac{11x+1}{1+4x^2(x+1)^{20}}]$

Explanation:

Solving for the derivative requires knowledge of the rule for the inverse tangent function:

$\frac{d}{dx}[\tan^{-1}(u)]=\frac{1}{1+u^2}\frac{du}{dx}$

In our case:

$u=2x(x+1)^{10}$

We can take the derivative of this using the product rule:

$\frac{du}{dx}=(2x)(10(x+1)^9(1))+(2)(x+1)^{10}=2(x+1)^9[10x+(x+1)]$

$\frac{du}{dx}=2(x+1)^9(11x+1)$

Now we can simply plug all of this into the above formula and we arrive at:

$\frac{dy}{dx}=[\frac{1}{1+(2x(x+1)^{10})^2}][2(x+1)^9(11x+1)]$

Simplifying this further gives:

$\frac{dy}{dx}=2(x+1)^{9}[\frac{11x+1}{1+4x^2(x+1)^{20}}]$

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Example Question #5 : Calculus Ab

f(x)=ln(x)x^2

What is the slope of f(x) at x=1 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

f(x)=g(x)h(x)

f'(x)=g'(x)h(x) + g(x)h'(x)

Following that procedure, we set ln(x) equal to g(x) and x^2 equal to h(x) .

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

which can be simplified to

f'(x)=x+2xln(x) = x(1+2ln(x)) .

Now plug in 1 to find the slope at x=1.

f'(1)= 1(1+2(0))=1

Remember that ln(1)=0 .

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Example Question #6 : Calculus Ab

Find the value of the derivate of the given function at the point (2,1) :

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

Possible Answers:

$\frac{3}5{}$

$\frac{1}2{}$

Correct answer:

$\frac{1}2{}$

Explanation:

To solve this problem, first, we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

$\\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point (2,1) . In this case, only the x value is important, so we evaluate our derivative at x=2 to get

$\\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\\ f'(2)=\frac{4+16-2}{36}\\f'(2)=\frac{18}{36}=\frac{1}{2}$

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Example Question #7 : Calculus Ab

Find the second derivative of the given function:

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

Possible Answers:

x^2+2x+1

$\frac{2x+2}{x}$

$\frac{2}{x^3}$

$\frac{x^2-1}{x^2}$

Correct answer:

$\frac{2}{x^3}$

Explanation:

To find the second derivative, first we need to find the first derivative. To find the first derivative we need to use the quotient rule as follows. So for the given function, we get the first derivative to be
$\\f'(x)=\frac{x\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2x+1)-(x^2+2x+1)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x)}{x^2}\\f'(x)=\frac{x\cdot (2x+2)-(x^2+2x+1)\cdot 1}{x^2}\\f'(x)=\frac{x^2-1}{x^2}$

Now we have to take the derivative of the derivative. To do this we need to use the quotient rule as shown below.

Thus, we get

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

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Example Question #8 : Calculus Ab

Find the derivative of the function

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

Possible Answers:

$\frac{1-\ln(x)^2}{[\ln(x)]^4}$

None of the other answers.

$\frac{1-\ln(x)}{[\ln(x)]^2}$

$\frac{\ln(x)-1}{[\ln(x)]^2}$

$\frac{(x^2)(1-\ln(x)^2)}{[\ln(x)]^4}$

Correct answer:

$\frac{\ln(x)-1}{[\ln(x)]^2}$

Explanation:

We find the answer using the quotient rule

$(\frac{f(x)}{g(x)})' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

and the product rule

(f(x)g(x))' = f(x)g'(x)+g(x)f'(x)

and then simplifying.

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

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

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

$=\frac{\ln(x)-1}{\ln(x)^2}$ or $\frac{\ln(x)-1}{[\ln(x)]^2}$ .

The extra brackets in the denominator are optional.

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Example Question #9 : Calculus Ab

If x^y = xy , find in terms of and .

Possible Answers:

$\frac{xy^2\ln(y)^2}{x\ln(x)-x^2\ln(x)^2}$

$\frac{xy\ln(y)^2}{\ln(x)-xy\ln(x)^2}$

None of the other answers

$\frac{xy\ln(y)^2}{\ln(x)-x^2y\ln(x)^2}$

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Correct answer:

$\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

Explanation:

Using a combination of logarithms, implicit differentiation, and a bit of algebra, we have

x^y=yx

$\ln(x^y)=\ln(xy)$

$y\ln(x)=\ln(x)+\ln(y)$

$y = \frac{\ln(y)}{\ln(x)}+1$

$y' = \frac{\ln(x)\frac{y'}{y}-\frac{\ln(y)}{x}}{\ln(x)^2}$ . Quotient Rule + implicit differentiation.

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

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

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

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

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

$y'=\frac{y\ln(y)}{x\ln(x)-xy\ln(x)^2}$

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Example Question #10 : Calculus Ab

Find the derivative of the function $y= \frac{\tan^{-1}(x)}{x}$

Possible Answers:

$\frac{x^2}{\tan^{-1}(x)\sqrt{1-x^2}}$

$\frac{x^2}{\tan(x)\sqrt{1-x^2}}$

None of the other answers

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

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

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{x-(1+x^2)\tan^{-1}(x)}{x^2(1+x^2)}$ .

Using the Quotient Rule and the fact $\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$ , we have

$y = \frac{\tan^{-1}(x)}{x}$

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

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

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

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

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

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Calculus AB : Determine Limits Using Algebraic Properties and the Squeeze Theorem

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Calculus Ab

Example Question #2 : Calculus Ab

$f(x)=x^2\tan^{-1}(3x^5)$

Example Question #3 : Calculus Ab

Example Question #4 : Calculus Ab

$y=\tan^{-1}(2x(x+1)^{10})$

Example Question #5 : Calculus Ab

Example Question #6 : Calculus Ab

Example Question #7 : Calculus Ab

Example Question #8 : Calculus Ab

Example Question #9 : Calculus Ab

Example Question #10 : Calculus Ab

All Calculus AB Resources

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