Specific Derivatives

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

$20x^{3}+\cos x$

$20x^{3}-\cos x$

$20x^3+4\cos x$

$5x^4\cos x +20x^3\sin x$

$20x^3+\frac{1}{\sin x}$

Explanation

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

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

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

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

$20x^{3}+\cos x$

$20x^{3}-\cos x$

$20x^3+4\cos x$

$5x^4\cos x +20x^3\sin x$

$20x^3+\frac{1}{\sin x}$

Explanation

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

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

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

Give the instantaneous rate of change of the function $f (x) = \left ( \frac{1}{3} \right ) ^{x}$ at .

$-\frac{ \ln 3}{27}$

$-27\ln 3$

$\frac{ \ln 3}{27}$

$\frac{ 1}{27}$

$-\frac{ 1}{27}$

Explanation

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}}{\mathrm{d} x}\left [ \left ( \frac{1}{3} \right ) ^{x }\right ] = \ln \frac{1}{3} \cdot \left ( \frac{1}{3} \right) ^{x }= -\frac{ \ln 3}{3^{x}}$

$f'(3) =-\frac{ \ln 3}{3^{3}} = -\frac{ \ln 3}{27}$

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

$60x^{2}-\sin x$

$40x^{2}-\sin x$

$40x^2-\cos x$

$120x+\sin x$

$x^5-\csc x$

Explanation

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

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

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

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

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

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

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

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

Give the instantaneous rate of change of the function $f (x) = \left ( \frac{1}{3} \right ) ^{x}$ at .

$-\frac{ \ln 3}{27}$

$-27\ln 3$

$\frac{ \ln 3}{27}$

$\frac{ 1}{27}$

$-\frac{ 1}{27}$

Explanation

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}}{\mathrm{d} x}\left [ \left ( \frac{1}{3} \right ) ^{x }\right ] = \ln \frac{1}{3} \cdot \left ( \frac{1}{3} \right) ^{x }= -\frac{ \ln 3}{3^{x}}$

$f'(3) =-\frac{ \ln 3}{3^{3}} = -\frac{ \ln 3}{27}$

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

$60x^{2}-\sin x$

$40x^{2}-\sin x$

$40x^2-\cos x$

$120x+\sin x$

$x^5-\csc x$

Explanation

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

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

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

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

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

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

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

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

Find the derivative of the following function:

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

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

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

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

Explanation

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

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

Find the derivative of the following function:

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

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

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

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

Explanation

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

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

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=?$

$\cos(x)$

$\sin(x)$

$\csc(x)$

$\sec(x)$

$-\sin(x)$

Explanation

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=?$

$\cos(x)$

$\sin(x)$

$\csc(x)$

$\sec(x)$

$-\sin(x)$

Explanation

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

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