AP Calculus BC - Rules of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, and Inverse Trigonometric

$f (x) = 7x^{2} + 6x - 9$

Give .

$f'(x) = 14x^{2}+6x$

Explanation

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Explanation

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

Differentiate $x^{999}$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{999}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 1,000x^{1,000}$

Explanation

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give .

$f''(x) =10.8x^{2} + 4.2x$

$f''(x) =10.8x^{2} + 4.2x - 0.5$

$f''(x) =10.8x^{2} + 8.4x$

$f''(x) =10.8x^{2} + 8.4x - 0.5$

$f''(x) =10.8x^{2} + 2.1x$

Explanation

First, find the derivative of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate to get .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

Find the derivative of:

$-\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

$-\frac{1+sin(x)\sqrt{1+x^2}}{\sqrt{1+x^2}}$

$\frac{1+cos(x)}{\sqrt{1+x^2}}$

$\frac{1-sin(x)}{\sqrt{1+x^2}}$

$\frac{cos(x)\sqrt{1-x^2}-sin(x)}{\sqrt{1-x^2}}$

Explanation

The derivative of inverse cosine is:

$\frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

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

Combine the two terms into one term.

$-\frac{1}{\sqrt{1-x^2}}-sin(x)$

$-\frac{1}{\sqrt{1-x^2}}-\frac{sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}= -\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

Find the derivative of the following function:

$f=\ln\sec(5x)$

$5\tan(5x)$

$\tan(5x)$

$5\sec(5x)\tan(5x)$

$\cot(5x)$

$5\tan(x)$

Explanation

The derivative of the function is equal to

$f'=5\tan(5x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

What is the rate of change of the function at the point ?

Explanation

The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is

$\frac{d}{dx}x^n=nx^{n-1}$ , and that the derivative of a constant is zero.

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

Therefore, the rate of change of f(x) at the point (1,6) is 14.

Find the derivative of the function $y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$

$x^2e^x\ln(x)\sin(x)$

Does not exist

None of the other answers

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

Explanation

To find the derivative of this function, we need to use the Fundemental Theorem of Calculus Part 1 (As opposed to the 2nd part, which is what's usually used to evaluate definite integrals)

$y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Start

$y'= \frac{d}{dx}\int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Take derivatives of both sides.

$y'= x^2e^x\ln(x)\sin(x)$ . "Cancel" the integral and the derivative. (Make sure that the upper bound on the integral is a function of , and that the lower bound is a constant before you cancel, otherwise you may need to use some manipulation of the bounds to make it so.)

Compute the derivative of the function

$y=\tan^3(2x)$ .

$6\tan^2(2x)\sec^2(2x)$

$3\tan^2(2x)\sec^2(2x)$

$2\tan^2(2x)\sec(2x)$

$6\tan^2(2x)$

$3\tan^2(2x)$

Explanation

Although written correctly by convention, the superscript that appears immediately after the trigonometric function $\tan$ may obscure the problem; the function

$y=\tan^3(2x)$ is equivalent to writing $y=\left (\tan(2x) \right )^3$ .

Using the fact that

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ ,

we apply the chain rule twice, using the power rule in the first step:

$y'=3\left (\tan(2x) \right )^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}\left (\tan(2x) \right )$

$=3\left (\tan(2x) \right )^2(\sec^2(2x)\cdot2)$

$=6\tan^2(2x)\sec^2(2x)$ .

(where in the last step, we have returned to the convention of writing the superscript immediately after $\tan$ )

Find the derivative of $y= e^{2}x$ .

$2e^{2x}$

None of the other answers

Explanation

This derivative uses the power rule. Keep in mind that the is not a part of the exponent of , and is thus being multiplied to . Since is a constant in front of , we have