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GRE Subject Test: Math : Calculus

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Example Questions

Example Question #11 : Finding Derivatives

Find the derivative of: $f(x)=\frac {2+x}{x}$

Possible Answers:

$\frac {2}{x}$

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

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

Correct answer:

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

Explanation:

Step 1: Define f(x),g(x) .

f(x)=x+2, g(x)=x

Step 2: Find f'(x),g'(x) .

f'(x)=1,g'(x)=1

Step 3: Plug in the functions/values into the formula for quotient rule: $\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$

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

The derivative of the expression is $-\frac {2}{x^2}$

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Example Question #1 : Quotient Rule

Find derivative $\left(\frac{f}{g}\right)'$ .

f(x)=5x^3

g(x)=e^x

Possible Answers:

$\left(\frac{f}{g}\right)'=\frac{3x^2-x^3}{e^x}$

$\left(\frac{f}{g}\right)'=\frac{15x^2+5x^3}{e^{2x}}$

$\left(\frac{f}{g}\right)'=\frac{15x^2+5x^3}{e^x}$

$\left(\frac{f}{g}\right)'=\frac{15x^2-5x^3}{e^{2x}}$

$\left(\frac{f}{g}\right)'=\frac{15x^2-5x^3}{e^x}$

Correct answer:

$\left(\frac{f}{g}\right)'=\frac{15x^2-5x^3}{e^x}$

Explanation:

This question yields to application of the quotient rule:

$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}$

So find and to start:

f(x)=5x^3

f'(x)=15x^2

g(x)=e^x

g'(x)=e^x

$\left(\frac{f}{g}\right)'=\frac{15x^2\cdot e^x-5x^3\cdot e^x}{(e^x)^2}$

$\left(\frac{f}{g}\right)'=\frac{e^x(15x^2-5x^3)}{(e^x)(e^x)}=\frac{15x^2-5x^3}{e^x}$

So our answer is:

$\frac{15x^2-5x^3}{e^x}$

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Example Question #5 : Quotient Rule

Find the second derivative of: $h(x)=\frac {x^3-2x}{x^5+4x^2}$

Possible Answers:

None of the Above

$\frac {6x^{11}+40x^9-47x^8-88x^6-140x^5-88x^3+256}{x^{15}+16x^{12}+96x^9+256x^6+256x^3}$

$\frac {-6x^{11}-40x^9+47x^8+88x^6+140x^5+88x^3-256}{x^{15}+16x^{12}+96x^9+256x^6+256x^3}$

$\frac {2x^5+8x^3+4x^2+8}{x^8+8x^5+16x^2}$

Correct answer:

None of the Above

Explanation:

Finding the First Derivative:

Step 1: Define f(x),g(x)

f(x)=x^3-2x, g(x)=x^5+4x^2

Step 2: Find f'(x), g'(x)

f'(x)=3x^2-2, g'(x)=5x^4+8x

Step 3: Plug in all equations into the quotient rule formula: $\frac {f'g-g'f}{g^2}$

$\frac {(3x^2-2)(x^5+4x^2)-[(5x^4+8x)(x^3-2x)]}{[(x^5+4x^2)]^2}$

Step 4: Simplify the fraction in step 3:

$\frac {3x^7+12x^4-2x^5-8x^2-[5x^7-10x^5+8x^4-16x^2]}{x^{10}+8x^7+16x^4}$

$=\frac {3x^7+12x^4-2x^5-8x^2-5x^7+10x^5-8x^4+16x^2}{x^{10}+8x^7+16x^4}$
$=\frac {-2x^7+8x^5+4x^4+8x^2}{x^{10}+8x^7+16x^4}$

Step 5: Factor an x^2 out from the numerator and denominator. Simplify the fraction..

$\frac {x^2(-2x^5+8x^3+4x^2+8)}{x^2(x^8+8x^5+16x^2)}=\frac {-2x^5+8x^3+4x^2+8}{x^8+8x^5+16x^2}$
We have found the first derivative..

Finding Second Derivative:

Step 6: Find f(x),g(x) from the first derivative function

f(x)=-2x^5+8x^3+4x^2+8, g(x)=x^8+8x^6+16x^2

Step 7: Find f'(x), g'(x)

f'(x)=-10x^4+24x^2+8x, g'(x)=8x^7+40x^4+32x

Step 8: Plug in the expressions into the quotient rule formula: $\frac {f'g-g'f}{g^2}$

$\frac {(-10x^4+24x^2+8x)(x^8+8x^5+16x^2)-[(8x^7+40x^4+32x)(-2x^5+8x^3+4x^2+8)]}{[(x^8+8x^6+16x^2)]^2}$

Step 9: Simplify:

$\frac {-10x^{12}-80x^9-160x^6+24x^{10}+192x^7+384x^4-...-64x^6-64x^{10}-...-256x}{[(x^8+8x^6+16x^2)]^2}$

I put "..." because the numerator is very long. I don't want to write all the terms...

Step 10: Combine like terms:

$\frac {-26x^{12}-88x^{10}-130x^9-296x^7-308x^6-680x^4-256x}{x^{16}+16x^{13}+96x^{10}+256x^7+256x^4}$

Step 11: Factor out and simplify:

Final Answer: $\frac {-26x^{11}-88x^9-130x^8-296x^6-308x^5-680x^3-256}{x^{15}+16x^{12}+96x^9+256x^6+256x^3}$ .

This is the second derivative.

The answer is None of the Above. The second derivative is not in the answers...

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Example Question #121 : Gre Subject Test: Math

Compute the derivative:

y=cos(sin(2x))

Possible Answers:

-2cos^2(2x)

-2cos(2x)sin(sin(2x))

-2sec(2x)tan(2x)

-2sin^2(2x)

-2cos(2x)sin^2(2x)

Correct answer:

-2cos(2x)sin(sin(2x))

Explanation:

This question requires application of multiple chain rules. There are 2 inner functions in y=cos(sin(2x)) , which are sin(2x) and .

The brackets are to identify the functions within the function where the chain rule must be applied.

Solve the derivative.

$\frac{dy}{dx}= -sin([sin(2x)]) \cdot (cos[2x]))\cdot 2=-2cos(2x)sin(sin(2x))$

The sine of sine of an angle cannot be combined to be sine squared.

Therefore, the answer is:

-2cos(2x)sin(sin(2x))

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Example Question #12 : Derivatives & Integrals

Find the derivative of the following function:

$h(x)=(3x^3-4x)^{33}$

Possible Answers:

$h'(x)=(297x^2+132)(3x^2-4x)^{33}$

$h'(x)=(297x^2-132)(3x^2-4x)^{32}$

$h'(x)=(297x^2-132)(3x^2-4x)^{33}$

$h'(x)=(9x^2-4)(3x^2-4x)^{32}$

$h'(x)=(297x^2)(3x^2-4x)^{32}$

Correct answer:

$h'(x)=(297x^2-132)(3x^2-4x)^{32}$

Explanation:

Recall chain rule for this problem

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

So if we are given the following,

$h(x)=(3x^3-4x)^{33}$

We can think of it like this

f(x)=3x^3-4x

f'(x)=9x^2-4

$g(x)=x^{33}$

$g'(x)=33x^{32}$

$(f(g(x)))'=(9x^2-4)(33)(3x^2-4x)^{32}$

Clean it up a bit to get:

$(f(g(x)))'=(297x^2-132)(3x^2-4x)^{32}$

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Example Question #1 : Chain Rule

What is the derivative of 3x^2*8y^3

Possible Answers:

3x^2*24y^2+8y^3*6x

6x*24y

3x^2*24y^2

8y^3*6x

3x*8y^2

Correct answer:

3x^2*24y^2+8y^3*6x

Explanation:

Chain Rule: Given: f(x)*g(y)

$\frac{\mathrm{dy} }{\mathrm{d} x}=f(x)g'(y)+g(y)f'(x)$

For this problem

f(x)=3x^2

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}3x^2=3(2)x^{2-1}=6x$

g(y)=8y^3

$g'(y)=\frac{\mathrm{d} }{\mathrm{d} x}8y^3=8(3)y^{3-1}=24y^2$

Plug the values into the Chain Rule formula and simplify:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^2*8y^3)=3x^2*24y^2+8y^3*6x$

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Example Question #1 : Defining Derivatives With Limits

Evaluate: $\lim _{x \to 2} \frac {x^2-4}{x^3-8}$

Possible Answers:

Does Not Exist

$-\frac {1}{3}$

$\frac {1}{3}$

Correct answer:

$\frac {1}{3}$

Explanation:

Step 1: Try plugging in x=2 into the denominator of the function. We want to make sure that the bottom does not become ...

2^3-8=8-8=0 .. We got zero, and we cannot have zero in the denominator. So, we must try and factor the function (numerator and denominator):

Step 2: Factor:

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

Step 3: Reduce:

$\frac {x+2}{x^2+2x+4}$

Step 4: Now that we got rid of the factor that made the denominator zero, we know that this function has a limit.

Step 5: Plug in x=2 into the reduced factor form:

Simplify as much as possible...

$\frac {2+2}{2^2+2(2)+4}=\frac {4}{12}=\frac {1}{3}$

The limit of this function as x approaches is $\frac {1}{3}$

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Example Question #1 : Implicit Differentiation

Differentiate the following with respect to .

$3y^2+12x-6\cos x=-4y^3$

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\sin x +2)}{y +2 y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\cos x +2)}{y +2 y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-\sin x +2}{y +2 y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\sin x -2}{y +2 y^2}$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\sin x +2)}{y +2 y^2}$

Explanation:

The first step is to differentiate both sides with respect to :

$3y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}+12x \ \frac{\mathrm{d} }{\mathrm{d} x}-6\cos x \ \frac{\mathrm{d} }{\mathrm{d} x}=-4y^3\frac{\mathrm{d} }{\mathrm{d} x}$

Note: Those that are functions of can be differentiated with respect to , just remember to mulitply it by $\frac{\mathrm{d} y}{\mathrm{d} x}$

$6y \ \frac{\mathrm{d}y }{\mathrm{d} x}+12+6\sin x =-12y^2\frac{\mathrm{d} y}{\mathrm{d} x}$

Now we can solve for $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$6y \ \frac{\mathrm{d}y }{\mathrm{d} x}+12 y^2\frac{\mathrm{d} y}{\mathrm{d} x}=-6\sin x -12$

$(6y +12 y^2) \ \frac{\mathrm{d} y}{\mathrm{d} x}=-6(\sin x +2)$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-6(\sin x +2)}{6(y +2 y^2)}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\sin x +2)}{y +2 y^2}$

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Example Question #2 : Implicit Differentiation

Find

$\frac{\mathrm{d} y}{\mathrm{d} x}$ for x^2+2y^3=8 .

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x^2}{2y^2}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x}{3y^2}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

Explanation:

Our first step would be to differentiate both sides with respect to :

$x^2 \frac{\mathrm{d} }{\mathrm{d} x}+2y^3 \frac{\mathrm{d} }{\mathrm{d} x}=8 \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of can be differentiated with respect to , just remember to multiply by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$2x + 6y^2 \frac{\mathrm{d}y }{\mathrm{d} x}=0$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-2x}{6y^2}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

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Example Question #3 : Implicit Differentiation

Differentiate the following to solve for $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$\cos y + 3x^3+4y=\sin x$

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\cos x-9x^2}{(\sin y -4 )}$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\cos x-9x^2}{(\sin y +4 )}$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{-\cos x+9x^2}{(\sin y +4y )}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$

Explanation:

Our first step is to differentiate both sides with respect to :

$\cos y \ \frac{\mathrm{d} }{\mathrm{d} x}+ 3x^3 \frac{\mathrm{d} }{\mathrm{d} x}+4y \frac{\mathrm{d} }{\mathrm{d} x}=\sin x \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of can by differentiated with respect to , just remember to multiply them by $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$-\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+ 9x^2 +4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x$

$-\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x-9x^2$

$(-\sin y +4 )\frac{\mathrm{d}y }{\mathrm{d} x}=(\cos x-9x^2)$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$

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GRE Subject Test: Math : Calculus

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #11 : Finding Derivatives

Example Question #1 : Quotient Rule

Example Question #5 : Quotient Rule

Example Question #121 : Gre Subject Test: Math

Example Question #12 : Derivatives & Integrals

Example Question #1 : Chain Rule

Example Question #1 : Defining Derivatives With Limits

Example Question #1 : Implicit Differentiation

Example Question #2 : Implicit Differentiation

Example Question #3 : Implicit Differentiation

All GRE Subject Test: Math Resources

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