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Calculus 1 : Differential Functions

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Example Questions

Example Question #231 : Differential Functions

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} \frac{x-xcosx}{sin^{2}x}$

Possible Answers:

$\pi^2$

- $-\frac{\pi}{2}$

DoesNotExist

Correct answer:

Explanation:

Multiplying by $\frac{x}{x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{x*(x-xcosx)}{x*sin^{2}x} = Lim _{x \rightarrow 0} \frac{x}{sinx}*\frac{x}{sinx}* \frac{1-cosx}{x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{x}{sinx}*\frac{x}{sinx}* \frac{1-cosx}{x} = Lim _{x \rightarrow 0} \frac{x}{sinx}*Lim _{x \rightarrow 0}\frac{x}{sinx}*Lim _{x \rightarrow 0} \frac{1-cosx}{x}$

And from the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$ and $Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

So:

$Lim _{x \rightarrow 0} \frac{x}{sinx}*Lim _{x \rightarrow 0}\frac{x}{sinx}*Lim _{x \rightarrow 0} \frac{1-cosx}{x}=1*1*0=0$

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Example Question #43 : Other Differential Functions

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} \frac{csc7x}{csc2x}$

Possible Answers:

Does Not Exist

$\frac{7}{2}$

$\frac{2}{7}$

$\frac{1}{14}$

Correct answer:

$\frac{2}{7}$

Explanation:

First observe that $\frac{csc7x}{csc2x} = \frac{sin2x}{sin7x}$

Multiplying by $\frac{2x}{2x}$ and $\frac{7x}{7x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{2x*7x*sin2x}{2x*7x*sin7x} = Lim _{x \rightarrow 0} \frac{sin2x}{2x}*\frac{7x}{sin7x}* \frac{2x}{7x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{sin2x}{2x}*\frac{7x}{sin7x}* \frac{2x}{7x}=Lim _{x \rightarrow 0} \frac{sin2x}{2x}*Lim _{x \rightarrow 0}\frac{7x}{sin7x}*Lim _{x \rightarrow 0} \frac{2x}{7x}$

And from the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

So:

$Lim _{x \rightarrow 0} \frac{sin2x}{2x}*Lim _{x \rightarrow 0}\frac{7x}{sin7x}*Lim _{x \rightarrow 0} \frac{2x}{7x}=1*1*\frac{2}{7}=\frac{2}{7}$

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Example Question #232 : Differential Functions

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} \frac{cos5x-1}{2x}$

Possible Answers:

$\frac{5}{2}$

$\frac{2\pi }{5}$

Does Not Exist

$\frac{2}{5}$

Correct answer:

Explanation:

Multiplying by $\frac{5x}{5x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{5x*(cos5x-1)}{5x*2x} = Lim _{x \rightarrow 0} \frac{5x}{2x}*\frac{cos5x-1}{5x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{5x}{2x}*\frac{cos5x-1}{5x}=Lim _{x \rightarrow 0} \frac{5x}{2x}*Lim _{x \rightarrow 0}\frac{cos5x-1}{5x}$

And from the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

So:

$Lim _{x \rightarrow 0} \frac{5x}{2x}*Lim _{x \rightarrow 0}\frac{cos5x-1}{5x}=\frac{5}{2}*0=0$

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Example Question #233 : Differential Functions

Given:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

$Lim _{x \rightarrow 0} \frac{1 - cosx}{x} = 0$

Evaluate the limit:

$Lim _{x \rightarrow 0} \frac{sin2x}{3x}$

Possible Answers:

$\frac{2\pi }{3}$

$\frac{3}{2}$

$\frac{2}{3}$

Does Not Exist

Correct answer:

$\frac{2}{3}$

Explanation:

Multiplying by $\frac{2x}{2x}$ we obtain:

$Lim _{x \rightarrow 0} \frac{2x*sin2x}{2x*3x} = Lim _{x \rightarrow 0} \frac{2x}{3x}*\frac{sin2x}{2x}$

Limit of product is the product of limits:

$Lim _{x \rightarrow 0} \frac{2x}{3x}*\frac{sin2x}{2x}=Lim _{x \rightarrow 0} \frac{2x}{3x}*Lim _{x \rightarrow 0}\frac{sin2x}{2x}$

And from the Pre-Question Text:

$Lim _{x \rightarrow 0} \frac{sinx}{x} = 1$

So:

$Lim _{x \rightarrow 0} \frac{2x}{3x}*Lim _{x \rightarrow 0}\frac{sin2x}{2x}=\frac{2}{3}*1=\frac{2}{3}$

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Example Question #41 : Other Differential Functions

Differentiate the function using known derivatives and applying the product, quotient, and chain rules.

f(x)=x*cos(2x)

Possible Answers:

f'(x)=cos(2x)+2*sin(2x)

f'(x)=cos(2x)+2x*sin(2x)

f(x)=-2sin(2x)

f'(x)=sin(2x)-2x*cos(2x)

f'(x)=cos(2x)-2x*sin(2x)

Correct answer:

f'(x)=cos(2x)-2x*sin(2x)

Explanation:

f(x)=x*cos(2x)

Using the product rule, $if \ f(x)=g(x) *h(x)$ ,

$then\ f'(x)=g'(x)*h(x) + g(x)*h'(x)$ ,

we observe the following:

$g(x)= x \ so\ g'(x)= 1$

$h(x) = cos(2x) \ so \ h'(x)=-2sin(2x) \ [This\ derivative \ uses\ the \ chain \ rule.]$

f'(x)= 1*cos(2x) + x*(-2sin(2x)) = cos(2x)-2xsin(2x)

f'(x)=cos(2x)-2xsin(2x) , which is our final answer.

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Example Question #46 : How To Find Differential Functions

Differentiate the function using known derivatives and applying the product, quotient, and chain rules.

$f(x)=sin^{2}(x) + cos^{2}(x)$

Possible Answers:

f'(x)=2sin(x) + 2cos(x)

f'(x)=4sin(x)*cos(x)

f'(x)=0

f'(x)=1

f'(x)=2cos(x) - 2sin(x)

Correct answer:

f'(x)=0

Explanation:

$f(x)=sin^{2}(x) + cos^{2}(x)$

We will need to use either the chain rule or product rule on both summands. The chain rule involves less simplifiction at the end. It states: $if \ f(x)=g(h(x))$ , $then\ f'(x)=g'(h(x))*h'(x)$ .

In this case, the outside function is, in fact, x^2 in both cases.

$g(x)= x^2 \ so\ g'(x)= 2x$

For the first summand:

$h(x) = sin(x) \ so \ h'(x)= cos(x)$

And for the the second:

$h(x) = cos(x) \ so \ h'(x)= -sin(x)$

f'(x)= 2sin(x)*cos(x) - 2cos(x)*sin(x) = 0

f'(x)=0 , which is our final answer.

[A function with a derivative of zero is a constant. You might recall the trig identity:

$sin^{2}(x) + cos^{2}(x) = 1$ , which is another way to solve this problem.]

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Example Question #1266 : Calculus

Differentiate the function using known derivatives and applying the product, quotient, and chain rules.

f(x)=(x+1)(x+2)(x+3)

Possible Answers:

f'(x)=1

f'(x)=3x+6

f'(x)=(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)

f'(x)=(x+2)(x+3) + 2(x+1)(x+3) + 3(x+1)(x+2)

f'(x)=x^3

Correct answer:

f'(x)=(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)

Explanation:

f(x)=(x+1)(x+2)(x+3)

You could expand this problem out, but let's evaluate the derivative using a modified product rule:

$if \ f(x)=g(x)*h(x)*j(x)$

$then\ f'(x)=g'(x)*h(x)*j(x)+g(x)*h'(x)*j(x)+g(x)*h(x)*j'(x)$ .

Setting each of the multiplicands as , , and , respectively, we get:

f'(x)=1*(x+2)(x+3) + (x+1)*1*(x+3) + (x+1)(x+2)*1

f'(x)=(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2) , which is our final answer.

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Example Question #51 : Other Differential Functions

Differentiate the function using known derivatives and applying the product, quotient, and chain rules.

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

Possible Answers:

$f'(x)=2*(3x^2 + 8x)^{2}$

$f'(x)=(x^{3}+4x^{2}+3)^{2}(3x^2 + 8x)$

$f'(x)=2x^{3}+8x^{2}+6$

f'(x)=2*(3x^2 + 8x)(6x+8)

$f'(x)=2*(x^{3}+4x^{2}+3)(3x^2 + 8x)$

Correct answer:

$f'(x)=2*(x^{3}+4x^{2}+3)(3x^2 + 8x)$

Explanation:

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

We evaluate this derivative using the chain rule:

$if \ f(x)=g(h(x))$ ,

$then\ f'(x)=g'(h(x))*h'(x)$ .

The outside function is:

$g(x) = x^2 \ so \ g'(x) = 2x$

The inside function is:

$h(x)=x^{3}+4x^{2}+3 \ so \ h'(x) = 3x^2 + 8x$

Therefore,

$f'(x)=2*(x^{3}+4x^{2}+3)(3x^2 + 8x)$ , which is our final answer.

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Example Question #51 : How To Find Differential Functions

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int tan(x)dx$

Possible Answers:

$\frac{sinx}{cosx} + c$

-ln(cosx)+c

$\frac{1}{1+x^2}+c$

$sec^{2}x + c$

ln(sinx)+c

Correct answer:

-ln(cosx)+c

Explanation:

We use the trig identity $tanx = \frac{sinx}{cosx}$ .

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

$Let \ u = cos(x), \ du=-sin(x)dx.$

$\int \frac{sin(x)}{cos(x)}dx = -\int \frac{du}{u}=-ln(u) +c= -ln(cosx)+c$ , which is our final answer.

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Example Question #1269 : Calculus

Solve the indefinite integral. If you cannot evaluate directly, use u-substitution.

$\int xe^{x^2}dx$

Possible Answers:

$\frac{1}{2}e^{x^{2}}$ + c

$2x^2e^{x^{2}}$ + c

$2e^{x^{2}}$ + c

$e^{2x}$ + c

$\frac{1}{2}e^{2x}$ + c

Correct answer:

$\frac{1}{2}e^{x^{2}}$ + c

Explanation:

$\int xe^{x^2}dx$

$Let \ u = x^2, \ du=2xdx$ .

$\int xe^{x^2}dx = \frac{1}{2}\int 2xe^{x^2}dx =\frac{1}{2}\int e^udu = \frac{1}{2}e^u = \frac{1}{2}e^{x^{2}}$ , which is our final answer.

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #231 : Differential Functions

Example Question #43 : Other Differential Functions

Example Question #232 : Differential Functions

Example Question #233 : Differential Functions

Example Question #41 : Other Differential Functions

Example Question #46 : How To Find Differential Functions

Example Question #1266 : Calculus

Example Question #51 : Other Differential Functions

Example Question #51 : How To Find Differential Functions

Example Question #1269 : Calculus

All Calculus 1 Resources

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