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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

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Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

z(y)=y^4tan(y)

Possible Answers:

$z'(y)=\frac{1}{y^2}[ysec^2(y)-4tan(y)]$

$z'(y)=-\frac{1}{y^2}[ysec^2(y)-4tan(y)]$

z'(y)=y^2[ysec^2(y)-4tan(y)]

$z'(y)=-\frac{1}{y^2}[ysec^2(y)+4tan(y)]$

z'(y)=y^3[ysec^2(y)+4tan(y)]

Correct answer:

z'(y)=y^3[ysec^2(y)+4tan(y)]

Explanation:

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STEPS

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Given:

z(y)=y^4tan(y)

Establish the derivative using the product rule:

z'(u*v)=u(v')+v(u')

z'(y)=y^4(sec^2(y))+4y^3(tan(x))

Distribute y terms

z'(y)=y^4sec^2(y)+4y^3tan(x)

Note and factor out the greatest common factor, y^3 to get the answer:

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ANSWER

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${\color{Red} z'(y)=y^3[ysec^2(y)+4tan(y)]}$

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Example Question #281 : Computation Of The Derivative

Find the derivative of the function:

s(r)=4sin(r)-3cos(r)

Possible Answers:

$s'(r)=-\frac{4cos(r)+sin(r)}{2}$

$s'(r)=\frac{3cos(r)+4sin(r)}{2}$

s'(r)=3cos(r)+4sin(r)

s'(r)=4cos(r)+3sin(r)

$s'(r)=-\frac{3cos(r)+4sin(r)}{2}$

Correct answer:

s'(r)=4cos(r)+3sin(r)

Explanation:

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STEPS

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Given:

s(r)=4sin(r)-3cos(r)

We can circumvent having to use the product rule by factoring out the constants before deriving the functions, like so:

$s'(r)=4{\color{Magenta} d(sin(r))}-3{\color{Teal} d(cos(r))}$

Now, we simply derive the trigonometric functions:

$s'(r)={\color{Magenta} 4cos(r)}-3{\color{Teal} (-sin(r))}$

Multiplying the negatives, we arrive at the correct answer:

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ANSWER

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${\color{Red} s'(r)=4cos(r)+3sin(r)}$

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Example Question #282 : Computation Of The Derivative

Find the derivative of the function:

x=csc(y)cot(y)

Possible Answers:

x'=-csc(y)(csc^2(y)+cot^2(y))

x'=2csc(y)(csc^2(y)+cot^2(y))

$x'=2csc(y)(csc^2(y)+\frac{1}{2}cot^2(y))$

x'=-csc(y)(csc^2(y)-cot^2(y))

x'=csc(y)(csc^2(y)+cot^2(y))

Correct answer:

x'=-csc(y)(csc^2(y)+cot^2(y))

Explanation:

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STEPS

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Given:

x=csc(y)cot(y)

Set up the product rule:

${\color{Magenta} d(u*v)=u(v')+v(u')}$

x'=csc(y)(-csc^2(y))+cot(y)(-csc(y)cot(y))

x'=-csc^3(y)-csc(y)cot^2(y)

Factor out a greatest common factor -csc(y), and we reach the answer:

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ANSWER

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${\color{Red} x'=-csc(y)(csc^2(y)+cot^2(y))}$

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Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

$s(r)=\frac{sin^2(r)}{cos^2(r)}$

Hint: Try to re-arrange the function first, and you can reach the answer faster and easier!

Possible Answers:

$s'(r)=\frac{1}{sin(r)sin(r)}$

s'(r)=-sin(r)sin(r)

s'(r)=-2tan(r)tan(r)

s'(r)=sin(r)sin(r)

s'(r)=2tan(r)sec^2(r)

Correct answer:

s'(r)=2tan(r)sec^2(r)

Explanation:

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STEPS

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Given:

$s(r)=\frac{sin^2(r)}{cos^2(r)}$

With a little bit of re-arranging the problem, we can circumvent the quotient rule entirely

$s(r)=\frac{(sin(r))^2}{(cos(r))^2}$

$s(r)=(\frac{sin(r)}{cos(r)})^{2}$

Substituting in tangent, we get

s(r)=tan(r)^2

We can now derive

s(r)=tan(r)^2

If we apply the general product rule, and remember to use the chain rule, we get:

$s'(r)=2tan(r)*{\color{Blue} (chain)}$

We can find our chained term by deriving tangent:

${\color{Blue} d(tan(r))=sec^2(r)}$

Plug this result into our chain and we arrive at the correct answer:

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ANSWER

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${\color{Red} s'(r)=2tan(r)sec^2(r)}$

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Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

$h(g)=e^{cos(g)}$

Possible Answers:

$h'(g)=2sin(g)e^{2sin(g)}$

$h'(g)=-sin(g)e^{cos(g)}$

$h'(g)=-2sin(g)e^{2sin(g)}$

$h'(g)=sin(g)e^{sin(g)}$

$h'(g)=sin(g)e^{cos(g)}$

Correct answer:

$h'(g)=-sin(g)e^{cos(g)}$

Explanation:

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STEPS

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Given:

$h(g)=e^{cos(g)}$

Understand the derivative of the exponential:

$h'(g)=e^{cos(g)}{\color{Blue} *(chain)}$

To obtain the chained term, we must derive the compound function element, in this case, the cos function contained in the exponent

${\color{Blue} d(cos(g))=-sin(g)}$

Plug the -sin back in:

$h'(g)=e^{cos(g)}{\color{Blue} *-sin(g)}$

Pull the sin to the front, and we arrive at the correct answer:

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ANSWER

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${\color{Red} h'(g)=-sin(g)e^{cos(g)}}$

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Example Question #285 : Computation Of The Derivative

Calculate the derivative of $\frac{ln(x)}{x}$ .

Possible Answers:

$\frac{1-ln(x)}{x}$

$\frac{-ln(x)}{x^2}$

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

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

Correct answer:

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

Explanation:

We don't have a formula for taking the derivative of this expression, so we'll have to use the quotient rule, since we have a fraction of functions.

The quotient rule is $(\frac{f}{g})'=\frac{g'f-f'g}{g^2}$ .

Applying it to $\frac{ln(x)}{x}$ , we get

$\frac{x*\tfrac{1}{x}-ln(x)*1}{x^2}=\frac{1-ln(x)}{x^2}$

Our answer is $\frac{1-ln(x)}{x^2}$ .

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Example Question #281 : Computation Of The Derivative

Find the derivative of $x^2\ln x$ .

Possible Answers:

$\ln x$

$x+x\ln x$

$x+\ln x$

$x+2x\ln x$

$2x\ln x$

Correct answer:

$x+2x\ln x$

Explanation:

Using the product rule, (fg)'=f*g'+g*f' , the derivative of $x^2\ln x$ is

$x^2(\tfrac{1}{x})+2x\ln x=x+2x\ln x$

Final answer:

$x+2x\ln x$

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Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of $15x^5-2x^4+11x^2+e^x-\ln x$ .

Possible Answers:

$25x^4+2x^3+11x+e^x+\ln x$

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

$15x^4-2x^3+11x+e^x-\ln x$

55x^4-6x^3+22x

$75x^4+8x^3-22x-e^x+\tfrac{1} {x}$

Correct answer:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

Explanation:

Take the derivative of each term.

$\tfrac{\mathrm{d} }{\mathrm{d} x}(15x^5)=75x^4$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-2x^4)=-8x^3$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(11x^2)=22x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(e^x)=e^x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-\ln x)=-\tfrac{1}{x}$

Add them:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

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Example Question #82 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of $2\sin x\cos x$ .

Possible Answers:

$2\sin2x$

$2\cos2x$

$-2\cos x\sin x$

$-\sin^2x+cos^2x$

Correct answer:

$2\cos2x$

Explanation:

There are two ways to solve this problem.

First, you can use a trig identity to replace $2\sin x\cos x$ with $\sin 2x$ . Using the chain rule, $(\sin 2x)'=2\cos2x$ .

Alternatively, you could use the product rule.

$2(\sin x(-\sin x)+\cos x(\cos x))=2(\cos^2x-\sin^2x)$

Since $\cos^2x-\sin^2x=\cos2x$ , our final answer is still $2\cos2x$ .

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Example Question #83 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of $\frac{e^x}{\sin x}$ .

Possible Answers:

$\frac{e^x\tan x-e^x}{\cos x}$

$\frac{e^x-e^x\tan x}{\sin x}$

$\frac{e^x\sin x-e^x\cos x}{\sin x}$

$\frac{e^x}{\cos x}$

$\frac{-e^x\tan x}{\sin x}$

Correct answer:

$\frac{e^x-e^x\tan x}{\sin x}$

Explanation:

For this problem, we need to use the quotient rule.

$(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$

$(\frac{e^x}{\sin x})'=\frac{\sin x(e^x)-e^x(\cos x)}{\sin^2x}$

Simplifying:

$e^x (\frac{\sin x-\cos x}{\sin^2x})=\frac{e^x-e^x\tan x}{\sin x}$

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AP Calculus AB : Derivative rules for sums, products, and quotients of functions

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #281 : Computation Of The Derivative

Example Question #282 : Computation Of The Derivative

Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #285 : Computation Of The Derivative

Example Question #281 : Computation Of The Derivative

Example Question #81 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #82 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #83 : Derivative Rules For Sums, Products, And Quotients Of Functions

All AP Calculus AB Resources

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