Derivatives & Integrals - GRE Subject Test: Math

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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

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

Answer

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

.. 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 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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Question

Solve for $\frac{\partial }{\partial x}$ :

Answer

To solve for the partial derivative, let all other variables be constants besides the variable that is derived with respect to.

In , the terms are constants.

Derive as accordingly by the differentiation rules.

$\frac{\partial }{\partial x}=s+2s^2x+sy$

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Question

Suppose the function $f=bcx+\frac{k}{c}+b^2ln(c)$ . Solve for $\frac{\partial f}{\partial c}$ .

Answer

Identify all the constants in function $f=bcx+\frac{k}{c}+b^2ln(c)$ .

Since we are solving for the partial differentiation of variable , all the other variables are constants. Solve each term by differentiation rules.

$\frac{\partial f}{\partial c}=bx-\frac{k}{c^2}+\frac{b^2}{c}$

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Question

Suppose the function $f=bcx+\frac{k}{c}+b^2ln(c)$ . Solve for $\frac{\partial f}{\partial c}$ .

Answer

Identify all the constants in function $f=bcx+\frac{k}{c}+b^2ln(c)$ .

Since we are solving for the partial differentiation of variable , all the other variables are constants. Solve each term by differentiation rules.

$\frac{\partial f}{\partial c}=bx-\frac{k}{c^2}+\frac{b^2}{c}$

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Question

Find

$\frac{\mathrm{d} y}{\mathrm{d} x}$ for .

Answer

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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Question

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

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

Answer

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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Question

Differentiate the following with respect to .

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

Answer

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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Question

Differentiate the following with respect to .

$\cos x +xy^2-\sin y=y^2$

Answer

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

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

Note: we can differentiate the terms that are functions of with respect to , just remember to multiply it by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

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

Note: The product rule was applied above:

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

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

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

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

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Question

Evaluate the following integral.

$\int x e^{4x}dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv-\int v\ du$

In this problem we have $\int x e^{4x}dx$ so we'll assign our substitutions:

and $dv=e^{4x} dx$

which means and $v=\frac{1}{4}e^{4x}$

Including our substitutions into the formula gives us:

$x\frac{1}{4}e^{4x}-\int \frac{1}{4}e^{4x}dx$

We can pull out the fraction from the integral in the second part:

$\frac{x}{4}e^{4x}-\frac{1}{4} \int e^{4x}dx$

Completing the integration gives us:

$\frac{x}{4}e^{4x}-\frac{1}{4} \left(\frac{1}{4}e^{4x}\right) +c$

$\frac{x}{4}e^{4x}-\frac{1}{16} e^{4x} +c$

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Question

Integrate the following.

$\int x \cos x \ dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv- \int v \ du$

So, our substitutions will be and $dv=\cos x \ dx$

which means and $v=\sin x$

Plugging our substitutions into the formula gives us:

$x\sin x-\int \sin x \ dx$

Since $\int \sin x \ dx=-\cos x + c$ , we have:

$x\sin x - (- \cos x + c)$ , or

$x\sin x + \cos x + c$

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Question

Evaluate the following integral.

$\int x^2 \ln x \ dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv-\int v \ du$

Our substitutions will be $u=\ln x$ and

which means $du= \frac{1}{x}dx$ and $v=\frac{1}{3}x^3$ .

Plugging our substitutions into the formula gives us:

$\frac{x^3}{3} \ln x -\int \frac{x^3}{3} \cdot \frac{1}{x} dx$

Look at the integral: we can pull out the $\frac{1}{3}$ and simplify the remaining $\frac{x^3}{x}$ as

$\frac{x^3}{3} \ln x -\frac{1}{3} \int x^2 dx$ .

We now solve the integral: $\int x^2 dx = \frac{1}{3} x^3 + c$ , so:

$\frac{x^3}{3} \ln x -\frac{1}{3} \left(\frac{1}{3}x^3 + c\right)$

$\frac{x^3}{3} \ln x - \frac{1}{9} x^3+c$

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Question

Evaluate the following integral.

$\int x \cos (2x+1)dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv - \int v \ du$ .

Our substitutions are and $dv= \cos (2x+1) dx$

which means and $v=\frac{1}{2} \sin (2x+1)$ .

Plugging in our substitutions into the formula gives us

$\frac{x}{2} \sin (2x+1)-\int \frac{1}{2} \sin (2x+1) dx$

We can pull $\frac{1}{2}$ outside of the integral.

$\frac{x}{2} \sin (2x+1)-\frac{1}{2} \int \sin (2x+1) dx$

Since $\int \sin (2x+1) dx =-\frac{1}{2} \cos (2x+1)+ c$ , we have

$\frac{x}{2} \sin (2x+1)- \frac{1}{2} \left(-\frac{1}{2} \cos (2x+1) + c\right)$

$\frac{x}{2} \sin (2x+1)+\frac{1}{4} \cos (2x+1) + c$

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Question

Find the local maximum for the function .

Answer

To find the local max, you must find the first derivative, which is .

Then. you need to set that equal to zero, so that you can find the critical points. The critical points are telling you where the slope is zero, and also clues you in to where the function is changing direction. When you set this derivative equal to zero and factor the function, you get , giving you two critical points at and .

Then, you set up a number line and test the regions in between those points. To the left of -1, pick a test value and plug it into the derivative. I chose -2 and got a negative value (you don't need the specific number, but rather, if it's negative or positive). In between -1 and 1, I chose 0 and got a possitive value. To the right of 1, I chose 2 and got a negative value. Then, I examine my number line to see where my function was going from positive to negative because that is what yields a maximum (think about a function going upwards and then changing direction downwards). That is happening at x=1.

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Question

Find the local maximum of the function.

Answer

When the derivative of a function is equal to zero, that means that the point is either a local maximum, local miniumum, or undefined. The derivative of is $nx^{n-1}$ . The derivative of the given function is

We must now set it equal to zero and factor.

Now we must plug in points to the left and right of the critical points to determine which is the local maximum.

This means the local maximum is at because the function is increasing at numbers less than -2 and decreasing at number between -2 and 6

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Question

Find the local maximum of the function.

Answer

The points where the derivative of a function are equal to 0 are called critical points. Critical points are either local maxs, local mins, or do not exist. The derivative of is $nx^{n-1}$ . The derivative of the function is

Now we must set it equal to 0 and factor to solve.

We must now plug in points to the left and right of the critical points into the derivative function to figure out which is the local max.

This means that the function is increasing until it hits x=-6, then it decreases until x=1, then it begins increasing again.

This means that x=-6 is the local max.

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Question

Find the coordinate of the local maximum of the folowing function.

Answer

At local maximums and minumims, the slope of the line tangent to the function is 0. To find the slope of the tangent line we must find the derivative of the function.

The derivative of is $nx^{n-1}$ . Thus the derivative of the function is

To find maximums and minumums we set it equal to 0.

So the critical points are at x=1 and x=2. To figure out the maximum we must plug each into the original function.

So the local max is at x=1.

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Question

Find the minimum of the function:

Answer

To find minimum take the derivative of the function and set it equal to zero.

$\frac{\mathrm{d} }{\mathrm{d} x}=24x+3$

Solve for x.

$x=\frac{-1}{8}$

Plugging x back in the equation will allow us to find the y value that results in the minimum.

$f(\frac{-1}{8})=12(\frac{-1}{8})^2+3(\frac{-1}{8})$

$f(\frac{-1}{8})=\frac{3}{16}+(\frac{-3}{8})$

$f(\frac{-1}{8})=\frac{-3}{16}$

The graph has a minimum at y=-3/16

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