Finding Derivatives - GRE Quantitative Reasoning

Card 0 of 72

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.

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

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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

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

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

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

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

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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:

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

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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 $ax, x\neq0$ , 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 $\frac{2x^2+4x+1}{(x+1)^2}$

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

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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...

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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:

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

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Question

Compute the derivative:

Answer

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:

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Question

Find the derivative of the following function:

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

Answer

Recall chain rule for this problem

So if we are given the following,

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

We can think of it like this

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

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

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

Clean it up a bit to get:

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

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Question

What is the derivative of

Answer

Chain Rule:

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

For this problem

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

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

Plug the values into the Chain Rule formula and simplify:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^28y^3)=3x^224y^2+8y^3*6x$

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Question

Compute the derivative:

Answer

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:

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Question

Find the derivative of the following function:

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

Answer

Recall chain rule for this problem

So if we are given the following,

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

We can think of it like this

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

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

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

Clean it up a bit to get:

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

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Question

What is the derivative of

Answer

Chain Rule:

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

For this problem

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

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

Plug the values into the Chain Rule formula and simplify:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^28y^3)=3x^224y^2+8y^3*6x$

Compare your answer with the correct one above

Question

Compute the derivative:

Answer

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:

Compare your answer with the correct one above

Question

Find the derivative of the following function:

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

Answer

Recall chain rule for this problem

So if we are given the following,

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

We can think of it like this

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

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

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

Clean it up a bit to get:

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

Compare your answer with the correct one above

Question

What is the derivative of

Answer

Chain Rule:

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

For this problem

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

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

Plug the values into the Chain Rule formula and simplify:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^28y^3)=3x^224y^2+8y^3*6x$

Compare your answer with the correct one above

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 $ax, x\neq0$ , 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 $\frac{2x^2+4x+1}{(x+1)^2}$

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