Quotient Rule - GRE Subject Test: Math

Card 0 of 7

Question

Find $\frac{dy}{dx}$ :

$y=\frac{ln(x)}{cos(x)}$

Answer

Write the quotient rule.

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

For the function $y=\frac{ln(x)}{cos(x)}$ , and , $f'(x)= \frac{1}{x}$ and .

Substitute and solve for the derivative.

$\frac{d}{dx}\left (\frac{f(x)}{g(x)} \right )=\frac{cos(x)(\frac{1}{x})-ln(x)(-sin(x))}{cos(x)^2}$

$\frac{d}{dx}\left (\frac{f(x)}{g(x)} \right )=\frac{\frac{cos(x)}{x}+ln(x)sin(x)}{cos(x)^2}$

Reduce the first term.

$y'=\frac{1}{xcos(x)}+\frac{ln(x)sin(x)}{cos(x)^2}$

Compare your answer with the correct one above

Question

Find the following derivative:

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

Given

Answer

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

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

Find the derivative of: $\frac{2x^2+2x+1}{x+1}$ .

Answer

Step 1: We need to define the quotient rule. The quotient rule says: $\frac {f'g-g'f}{g^2}$ , where is the derivative of and is the derivative of

Step 2: We need to review how to take derivatives of different kinds of terms. When we are taking the derivative of terms in a polynomial, we need to follow these rules:

Rule 1: For any term with an exponent, the derivative of that term says: Drop the exponent and multiply it to the coefficient of that term. The new exponent of the derivative is lower than the previous exponent.

Example: $6x^2->6(2)x^{2-1}=12x^1=12x$

Rule 2: For any term in the form , the derivative of that term is just , the coefficient of that term.

Ecample:

Rule 3: The derivative of any constant is always

Step 3: Find and :

Step 4: Plug in all equations into the quotient rule:

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

Step 5: Simplify the fraction in step 4:

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

Step 6: Combine terms in the numerator in step 5:

$\frac{2x^2+4x+1}{(x+1)^2}$ .

The derivative of $\frac{2x^2+2x+1}{x+1}$ is $\frac{2x^2+4x+1}{(x+1)^2}$

Compare your answer with the correct one above

Question

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

Answer

Step 1: Define .

Step 2: Find .

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

Compare your answer with the correct one above

Question

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

Answer

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...

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Tap the card to reveal the answer