Finding Derivatives

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GRE Quantitative Reasoning › Finding Derivatives

Questions 1 - 10

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

None of the Above

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

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

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

Explanation

Finding the First Derivative:

Step 1: Define

Step 2: Find

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 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 from the first derivative function

Step 7: Find

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...

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

None of the Above

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

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

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

Explanation

Finding the First Derivative:

Step 1: Define

Step 2: Find

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 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 from the first derivative function

Step 7: Find

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...

Compute the derivative:

Explanation

This question requires application of multiple chain rules. There are 2 inner functions in , which are 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:

Compute the derivative:

Explanation

This question requires application of multiple chain rules. There are 2 inner functions in , which are 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:

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

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

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

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

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

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

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:

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

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

So our answer is:

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

Find the following derivative:

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

Given

$\left(\frac{f}{g}\right)'=\frac{78x-49}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78x-49}{x^{16}}$

$\left(\frac{f}{g}\right)'=\frac{-78x+49}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78x-343}{x^{18}}$

Explanation

This question asks us to find the derivative of a quotient. Use the quotient rule:

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

Start by finding and .

So we get:

$\left(\frac{f}{g}\right)'=\frac{(-39x^2+14x)(x^9)-(-13x^3+7x^2)(9x^8)}{(x^9)^2}$

Whew, let's simplify

$\left(\frac{f}{g}\right)'=\frac{(-39x^{11}+14x^{10})-(-117x^{11}+63x^{10})}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{-39x^{11}+14x^{10}+117x^{11}-63x^{10}}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{78x^{11}-49x^{10}}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{78x-49}{x^{8}}$

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

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

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

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

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

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

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:

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

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

So our answer is:

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

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

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

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

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

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

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

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:

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

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

So our answer is:

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

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

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

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

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

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

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

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:

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

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

So our answer is:

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

Find the following derivative:

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

Given

$\left(\frac{f}{g}\right)'=\frac{78x-49}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78x-49}{x^{16}}$

$\left(\frac{f}{g}\right)'=\frac{-78x+49}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78}{x^{8}}$

$\left(\frac{f}{g}\right)'=\frac{-78x-343}{x^{18}}$

Explanation

This question asks us to find the derivative of a quotient. Use the quotient rule:

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

Start by finding and .

So we get:

$\left(\frac{f}{g}\right)'=\frac{(-39x^2+14x)(x^9)-(-13x^3+7x^2)(9x^8)}{(x^9)^2}$

Whew, let's simplify

$\left(\frac{f}{g}\right)'=\frac{(-39x^{11}+14x^{10})-(-117x^{11}+63x^{10})}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{-39x^{11}+14x^{10}+117x^{11}-63x^{10}}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{78x^{11}-49x^{10}}{x^{18}}$

$\left(\frac{f}{g}\right)'=\frac{78x-49}{x^{8}}$

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