Finding Derivatives - Math

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Question

Find the derivative of the following function:

$f(x) = x^3 + \frac{1}{x}$

Answer

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

is simply

We can rewrite $\frac{1}{x}$ as

$x^{-1}$

and using the power rule again, we get a derivative of

$-x^{-2}$ or $-\frac{1}{x^2}$

So the answer is

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

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Question

Which of the following best represents $\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ ?

Answer

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$

the bottom function is and the top function is . This makes the bottom derivative and the top derivative .

Substituting these into the Quotient Rule formula resulting in the following.

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$

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Question

What is $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} ?$

Answer

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} =f(x)g'(x)+f'(x)g(x)$ .

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Question

What is the first derivative of $5x^4+\sin x$ ?

Answer

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(4*5x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

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Question

What is the second derivative of $5x^4+\sin x$ ?

Answer

To find the second derivative, we need to start by finding the first one.

Since we're adding terms, we take the derivative of each part separately. For , we can use the power rule, which states that we multiply the variable by the current exponent and then lower the exponent by one. For sine, we use our trigonometric derivative rules.

Remember, $\frac{\mathrm{d} }{\mathrm{d} x} \sin x=\cos x$ .

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(45x^{4-1})+\cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}5x^4+\sin x=(20x^{3})+\cos x$

Now we repeat the process, but using $20x^3+\cos x$ as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x} \cos x=-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(320x^{3-1})-\sin x$

$\frac{\mathrm{d} }{\mathrm{d} x}20x^3+\cos x=(60x^{2})-\sin x$

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Question

An ellipse is represented by the following equation:

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

What is the slope of the curve at the point (3,2)?

Answer

It would be difficult to differentiate this equation by isolating . Luckily, we don't have to. Use $\small \frac{dy}{dx}$ to represent the derivative of with respect to and follow the chain rule.

(Remember, $\small \frac{dx}{dx}$ is the derivative of with respect to , although it usually doesn't get written out because it is equal to 1. We'll write it out this time so you can see how implicit differentiation works.)

$\small 8x^{2}+3y^{2}+6x+y-104 = 0$

$\small 16x\frac{dx}{dx}+6y\frac{dy}{dx}+6\frac{dx}{dx}+\frac{dy}{dx}=0$

$\small 16x(1)+6y\frac{dy}{dx}+6(1)+\frac{dy}{dx}=0$

$\small 16x+6y\frac{dy}{dx}+6+\frac{dy}{dx}=0$

Now we need to isolate $\small \frac{dy}{dx}$ by first putting all of these terms on the same side:

$\small \small 6y\frac{dy}{dx}+\frac{dy}{dx}=-16x-6$

$\small \frac{dy}{dx}(6y+1)=-16x-6$

$\small \frac{dy}{dx}=\frac{-16x-6}{6y+1}$

This is the equation for the derivative at any point on the curve. By substituting in (3, 2) from the original question, we can find the slope at that particular point:

$\small \frac{dy}{dx}=\frac{-16(3)-6}{6(2)+1}=\frac{-48-6}{12+1}=\frac{-54}{13}$

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Question

If $f(x)=\frac{2}{3}x^3+7x^2-12x$ , what is ?

Answer

For this problem, we can use the power rule. The power rule states that we multiply each variable by its current exponent and then lower that exponent by one.

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)=(3\frac{2}{3}x^{3-1})+(27x^{2-1})-(1*12x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)=(2x^{2})+(14x^{1})-(12x^{0})$

Anything to the zero power is one, so .

Therefore, $f'(x)=2x^{2}+14x-12$ .

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Question

Find the derivative of the following function:

Answer

We use the power rule on each term of the function.

The first term

becomes

$2 \times 6 x^2 = 12 x$ .

The second term

becomes

.

The final term, 7, is a constant, so its derivative is simply zero.

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Question

Give the average rate of change of the function $f(x) = 4 ^{x}$ on the interval .

Answer

The average rate of change of on interval is

$\frac{f (b ) - f(a)}{b-a}$

Substitute:

$\frac{f (4 ) - f(3)}{4-3} = \frac{4^{4} - 4^{3}}{1} = \frac{256 - 64}{1} = 192$

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Question

What is the derivative of ?

Answer

To get $\frac{\mathrm{d} }{\mathrm{d} x}(2x)$ , we can use the power rule.

Since the exponent of the is , as , we lower the exponent by one and then multiply the coefficient by that original exponent:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=12x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2x^{0}$

Anything to the power is .

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=21$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x)=2$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2x^{2-1})+(15x^{1-0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})+(08x^{0-1})$

Notice that $(08x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=51$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5$

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Question

What is the derivative of ?

Answer

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as , as anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})+(05x^{0-1})$

Notice that $(05x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3x^{3-1})+(12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2$

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Question

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=?$

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})+(08x^{0-1})$

Notice that $08x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(26x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})+(0x^{0-1})$

Notice that $(0x^{0-1})=0$ since anything times zero is zero.

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1x^{1-1})$ .

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

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Question

What is the derivative of ?

Answer

To solve this equation, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})+(04x^{0-1})$

Notice that $(04x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{2-1})+(113x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(212x^{1})+(113x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

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Question

What is the derivative of $2x^4+\frac{1}{2}x$ ?

Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{4-1})+(1\frac{1}{2}x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(42x^{3})+(\frac{1}{2}x^{0})$

Remember that anything to the zero power is equal to one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=(8x^{3})+(\frac{1}{2}1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x^4+\frac{1}{2}x)=8x^{3}+\frac{1}{2}$

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Question

What is the derivative of ?

Answer

To take the derivative of this equation, we can use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent.

We are going to treat as since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})+(012x^{0-1})$

Notice that $(012x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{1-1})$

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=(3x^{0})$

As stated before, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x+12)=3$

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Question

What is the derivative of ?

Answer

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{2-1})+(112x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(25x^{1})+(112x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

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Question

What is the derivative of ?

Answer

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

For this problem that would look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{2-1})+(015x^{0-1})$

Notice that $(015x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

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