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

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Question

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give .

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Answer

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

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Question

Find $\frac{dy}{dx}$ if $x^3 -x\ln y=ye^x$ .

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Answer

This function is implicit, because y is not defined directly in terms of only x. We could try to solve for y, but that would be difficult, if not impossible. The easier solution would be to employ implicit differentiation. Our strategy will be to differentiate the left and right sides by x, apply the rules of differentiation (such as Chain and Product Rules), group dy/dx terms, and solve for dy/dx in terms of both x and y.

$x^3-x\ln y=ye^{x}$

$\frac{d}{dx}[x^3-x\ln y]=\frac{d}{dx}[ye^x]$

$\frac{d}{dx}[x^3]-\frac{d}{dx}[x\ln y]=\frac{d}{dx}[ye^x]$

We will need to apply both the Product Rule and Chain Rule to both the xlny and the terms.

According to the Product Rule, if f(x) and g(x) are functions, then $\frac{d}{dx}[f(x)g(x)]=f(x)\cdot \frac{d}{dx}[g(x)]+g(x)\cdot \frac{d}{dx}[f(x)]$ .

And according to the Chain Rule,

$\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)$

$3x^2-(x\cdot \frac{d}{dx}[\ln y]+\ln y\cdot \frac{d}{dx}[x])=e^x\cdot \frac{d}{dx}[y]+y\cdot \frac{d}{dx}[e^x]$

$3x^2-(x\cdot \frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{d}{dy}[y]\cdot \frac{dy}{dx}+y\cdot e^x$

$3x^2-(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{dy}{dx}+ye^x$

$3x^{2}-\frac{x}y{}\cdot \frac{dy}{dx}-\ln y=e^x\cdot \frac{dy}{dx}+ye^x$

Now we will group the dy/dx terms and move everything else to the opposite side.

$3x^2-\ln y-ye^x=e^x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=(e^x+\frac{x}{y})\frac{dy}{dx}$

Then, we can solve for dy/dx.

$\frac{3x^2-\ln y-ye^x}{e^x+\frac{x}{y}} =\frac{dy}{dx}$

To remove the compound fraction, we can multiply the top and bottom of the fraction by y.

$\frac{y(3x^2-\ln y-ye^x)}{y(e^x+\frac{x}{y})} =\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

The (ugly) answer is $\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$ .

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Question

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

Give .

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Answer

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

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

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

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

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Question

Differentiate $x^{999}$ .

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Answer

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

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

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Question

Give the second derivative of $x^{777}$ .

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Answer

Find the derivative of $x^{777}$ , then find the derivative of that expression.

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

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

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Question

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

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Answer

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 .

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

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

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Question

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

Give .

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Answer

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 .

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

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

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

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

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Question

Find the derivative of:

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Answer

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.

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

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

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Question

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

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Answer

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.

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Question

Find the derivative of the function $y = \int_{0}^{x^2} e^{-t^2}dt$

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Answer

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$y = \int_{0}^{x^2} e^{-t^2}dt$ . Start

$y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$ . Take the derivative of both sides with respect to .

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's in this case), and that the upper limit of the integral is a function of , (it's in this case).

Afterward, plug in for , and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$ .

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Question

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

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Answer

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.)

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Question

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

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Answer

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

$y' = \frac{d}{dx}(e^2x) = e^2\frac{d}{dx}(x) = e^2 \times 1 = e^2$ .

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Question

If $f(x) = 3^{2x}$ , find .

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Answer

In general, the derivative for an exponential function is

$\ln(a) \times a^u \times \frac{du}{dx}$ .

In our case , , so we have

$f'(x) = \ln(3) \times 3^{2x} \times (2x)' = 2\ln(3) \times 3^{2x}$ .

Hence

$f'(1) = 2\ln(3) \times 3 = 6\ln{(3)}$ .

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Question

Find the derivative of the following function:

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

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Answer

The derivative of the function is equal to

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

and was found using the following rules:

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

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Question

Find the derivative of the following function:

$f(x)=\log_3(x^2)$

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Answer

The derivative of the function is equal to

$f'(x)=\frac{2}{x\ln(3)}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \log_a(u)=\frac{1}{u\ln(a)}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Question

Given h(x), find h'(x).

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Answer

Given h(x), find h'(x).

We need to derive a function composed of trigonometric terms.

Let's recall the rules

Derivative of sine is cosine

Derivative of cosine is negative sine

Derivative of tangent is secant squared.

Put this all together to get:

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Question

Find the first derivative of the given function.

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Answer

Find the first derivative of the given function.

So, here we need to derive a function with trigonometric terms. Let's recall the rules

The derivative of secant is secant times tangent

The derivative of cotangent is negative cosecant squared.

The derivative of any constant term is 0

Put all this together to get:

And simplify to get:

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Question

Compute the derivative of the function

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

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Answer

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 $\tan$ )

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Question

Compute the derivative of the function

$y=\ln(1+e^{-4x})$ .

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Answer

Use the chain rule: with the outer function as $f(x)=\ln(x)$ and the inner function as $g(x)=1+e^{-4x}$ .

We have then:

$y'=\left (\frac{1}{1+e^{-4x}} \right )(e^{-4x}\cdot-4)$

(where we have used the chain rule again to compute the derivative of the inner function )

We can simplify this further (into the format of the answer choices) as follows:

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

$=\frac{-4e^{-4x}}{1+e^{-4x}}\cdot\frac{e^{4x}}{e^{4x}}$

(multiplication by a convenient form of 1)

$=\frac{-4}{e^{4x}+1}$ .

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Question

Find $\frac{dy}{dx}$ if $x^3 -x\ln y=ye^x$ .

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Answer

This function is implicit, because y is not defined directly in terms of only x. We could try to solve for y, but that would be difficult, if not impossible. The easier solution would be to employ implicit differentiation. Our strategy will be to differentiate the left and right sides by x, apply the rules of differentiation (such as Chain and Product Rules), group dy/dx terms, and solve for dy/dx in terms of both x and y.

$x^3-x\ln y=ye^{x}$

$\frac{d}{dx}[x^3-x\ln y]=\frac{d}{dx}[ye^x]$

$\frac{d}{dx}[x^3]-\frac{d}{dx}[x\ln y]=\frac{d}{dx}[ye^x]$

We will need to apply both the Product Rule and Chain Rule to both the xlny and the terms.

According to the Product Rule, if f(x) and g(x) are functions, then $\frac{d}{dx}[f(x)g(x)]=f(x)\cdot \frac{d}{dx}[g(x)]+g(x)\cdot \frac{d}{dx}[f(x)]$ .

And according to the Chain Rule,

$\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)$

$3x^2-(x\cdot \frac{d}{dx}[\ln y]+\ln y\cdot \frac{d}{dx}[x])=e^x\cdot \frac{d}{dx}[y]+y\cdot \frac{d}{dx}[e^x]$

$3x^2-(x\cdot \frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{d}{dy}[y]\cdot \frac{dy}{dx}+y\cdot e^x$

$3x^2-(\frac{x}{y}\cdot \frac{dy}{dx}+\ln y)=e^x\cdot \frac{dy}{dx}+ye^x$

$3x^{2}-\frac{x}y{}\cdot \frac{dy}{dx}-\ln y=e^x\cdot \frac{dy}{dx}+ye^x$

Now we will group the dy/dx terms and move everything else to the opposite side.

$3x^2-\ln y-ye^x=e^x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}=(e^x+\frac{x}{y})\frac{dy}{dx}$

Then, we can solve for dy/dx.

$\frac{3x^2-\ln y-ye^x}{e^x+\frac{x}{y}} =\frac{dy}{dx}$

To remove the compound fraction, we can multiply the top and bottom of the fraction by y.

$\frac{y(3x^2-\ln y-ye^x)}{y(e^x+\frac{x}{y})} =\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$

The (ugly) answer is $\frac{dy}{dx}=\frac{(3x^2y-y\ln y-y^2e^x)}{e^xy+x}$ .

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