Derivatives of functions - AP Calculus AB

Card 1 of 1034

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

Find the derivative of the following:

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

Tap to reveal answer

Answer

The easiest way to approach this problem is to break it up into terms to get:

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

This simplifies to:

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

This then becomes a simple derivative in which you get:

$\small f'(x)=0+\frac{0-2x}{x^4}$

Which after simplifying you get:

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

← Didn't Know|Knew It →

Question

Find the derivative:

$\small x^4+x-cos(x)$

Tap to reveal answer

Answer

This is a derivative of sums with a trigonometric function thrown in there.

Upon simplifying you get:

$\small f'(x)=\small 4x^3+1+sin(x)$

Keeping in mind the derivative of cos(x) is -sin(x)

← Didn't Know|Knew It →

Question

Find the derivative.

$\sin (x)$

Tap to reveal answer

Answer

The derivative of $\sin (x)$ is $\cos (x)$ . (Memorization)

← Didn't Know|Knew It →

Question

Calculate the derivative:

$\small \small \small f(x)=sin(x)(sec(x))$

Tap to reveal answer

Answer

This is a chain rule using trigonometric functions.

$\small \small f'(x)=cos(x)sec(x)+sec(x)tan(x)sin(x)$

Which upon simplifying is:

$\small \small \small f'(x)=1+tan^2(x)$

← Didn't Know|Knew It →

Question

Find the derivative.

$\sin (2x)$

Tap to reveal answer

Answer

Use the chain rule to find the derivative: $\frac{d}{dx}f(g(x))=g'(x)f'(g(x))$

Thus, $\frac{d}{dx}\sin (2x)=2\cos (2x)$ .

← Didn't Know|Knew It →

Question

Find the derivative.

$12x^{12}-13x^2$

Tap to reveal answer

Answer

Use the power rule to find the derivative.

$\frac{d}{dx}12x^{12}=144x^{11}$

$\frac{d}{dx}13x^2=26x$

Thus, the derivative is $144x^{11}-26x$

← Didn't Know|Knew It →

Question

Find the derivative.

Tap to reveal answer

Answer

Use the power rule to find the derivative.

$\frac{d}{dx}7x^2=14x$

$\frac{d}{dx}4x=4$

The derivative of a constant is zero.

Thus, the derivative is .

← Didn't Know|Knew It →

Question

Use the method of your choice to find the derivative.

Tap to reveal answer

Answer

The easiest way to find this derivative is to FOIL, and then use the power rule.

$\frac{d}{dx}2x^2+18x+40=4x+18$

← Didn't Know|Knew It →

Question

Find the derivative.

$\sin (x)\cos (x)$

Tap to reveal answer

Answer

Use the product rule to find this derivative.

$\frac{d}{dx}\sin(x)\cos (x)=-\sin^2(x)+\cos^2(x)$

← Didn't Know|Knew It →

Question

Find :

Tap to reveal answer

Answer

This is the product rule, which is: (derivative of the first)(second)+(derivative of the second)(first)

So:

← Didn't Know|Knew It →

Question

If $f(x)=\sin x$ then $f'\left({\frac{\pi }{4}}\right)=$

Tap to reveal answer

Answer

To calculate the derivative of this function at the desired point, first recall that,

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

$f(x)=\sin x$

$f^{}(x)=\cos x$

Now, substitute the value into the derivative function to solve.

$f'(\frac{\pi }{4}) = cos\frac{\pi }{4} = cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

← Didn't Know|Knew It →

Question

If the position of a particle over time is represented by $p=t^{3}-16t^{2}-7t$ then what is the particle's instantaneous acceleration at ?

Tap to reveal answer

Answer

The answer is .

Since velocity is the first derivative of the position function, take the derivative once. Then, recall that the acceleration function is the second derivative of position thus the derivative needs to be taken one more time.

$p=t^{3}-16t^{2}-7t$

$velocity =p'=3t^{2}-32t-7$

$acceleration =p''=6t-32$

$p''(4)=6(4)-32 = -8$

← Didn't Know|Knew It →

Question

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when ?

Tap to reveal answer

Answer

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .

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

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

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

$f'(2)=2(2)-2(2)(1-2^2)^{-2}$

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

← Didn't Know|Knew It →

Question

Consider the function:

$f(x) = -6e^{-x} + x$

The relative minimum for this function is at:

Tap to reveal answer

Answer

To find any relative minimum, one first needs to find the critical points by setting the first derivative equal to zero:

$f'(x) = 6e^{-x} + 1 = 0$

However, the first derivative is positive for all real values of x, since the exponential function is always positive. Thus, there are no values for which , and therefore no critical points and no relative minimum.

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

$x^{2}+12x-36$

Tap to reveal answer

Answer

Using the power rule which states,

$\frac{d}{dx}x^n=nx^{n-1}$

you can move the from $x^{2}$ to the front and decrease the exponent by which makes it .

For , any term that has an exponent of , the coefficient is its derivative.

Thus, the derivative of is .

Since does not have a variable attached, the derivative will be .

Add your derivatives to get .

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

$x^{3}+8x^{2}-16x+9$

Tap to reveal answer

Answer

Use the power rule to move the exponent of each term to the front, and multiply it with the existing coefficient to create the new coefficient for the derivative.

In mathematical terms, the power rule states,

$\frac{d}{dx}x^n=nx^{n-1}$

Applying the power rule to the first term creates $3x^{2}$ .

Next, move the from $8x^{2}$ to the front and multiply it by , and decrease the exponent by 1 to get .

Next, since does not have an exponent, the derivative of that will be .

Lastly, has a derivative of because there is not variable attached to it.

Therefore the derivative becomes,

$3x^{2}+16x-16$ .

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

$\sqrt{x}$

Tap to reveal answer

Answer

To find the derivative, use the power rule.

In mathematical terms, the power rule states,

$\frac{d}{dx}x^n=nx^{n-1}$

$\sqrt{x}$ is the same as $x^{\frac{1}{2}}$ .

Therefore, move the exponent to the front, and then decrease it by one to get

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

After simplifying, you get

$\frac{1}{2\sqrt{x}}$ .

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

$\left ( x^2-4\right )$

Tap to reveal answer

Answer

Having a binomial does not change the rules for the power rule. You still move the exponent to the front, and decrease the exponent by .

In mathematical terms, the power rule states,

$\frac{d}{dx}x^n=nx^{n-1}$

Constants still have a derivative of

Thus, giving you a final answer of

$\f(x)=x^2-4 \f'(x)=2x^{2-1}-0 \f'(x)=2x$

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

$\left ( x^2-2\right )^2$

Tap to reveal answer

Answer

Use the chain rule to move the exponent of the binomial to the front, and decrease the exponent by 1. Next, take the derivative of what is on the inside and multiply it with what is one the outside.

In mathematical terms the chain rule is,

$\frac{d}{dx}(f(x))^n\rightarrow n(f(x))\cdot f'(x)$

Identify f(x) and its derivative first.

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

Substituting the function and its derivative into the chain rule formula, the final derivative becomes

$\f(x)=(x^2-2)^2 \f'(x)=2(x^2-2)(2x) \f'(x)=4x(x^2-2)$

Thus, giving you an answer of $4x\left ( x^2-2\right )$ .

← Didn't Know|Knew It →

Question

Calculate the derivative of the following:

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

Tap to reveal answer

Answer

To find the derivative, use the quotient rule.

The quotient rule requires you to do the following:

$\frac{(Denominator)(Deriv. Numerator)-(Numerator)(Deriv. Denominator)}{(Denominator^2)}$

When you apply it to this problem, you get a final answer of,

$\\text{Numerator}=x^2 \\text{Deriv. of Numerator}=2x \\text{Denominator}=2 \\text{Deriv. of Denominator}=0$

$\f'(x)=\frac{2(2x)-x^2(0)}{2^2} \\f'(x)=\frac{4x}{4} \\f'(x)=x$

← Didn't Know|Knew It →

0

Knew It