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

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Example Questions

Example Question #71 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the first derivative of the function:

$t=\frac{s^2+5s-1}{s^2}$

Possible Answers:

$t'=-\frac{s^4}{15s^2-2s}$

$t'=-\frac{5s^2-8s}{s^4}$

$t'=\frac{-5s+2}{s^3}$

$2(\frac{s^4}{15s^2-2s})$

$t'=\frac{s^4}{15s^2-2s}$

Correct answer:

$t'=\frac{-5s+2}{s^3}$

Explanation:

We are given the function:

$t=\frac{s^2+5s-1}{s^2}$

As the function given to us is a quotient, we must use the quotient rule for derivatives to get the derivative of the function.

$t'=\frac{[(s^2)(2s+5)]-[(s^2+5s-1)(2s)]}{s^4}$

Technically, this derivative is correct, but it is not at all optimal. Let's clean things up and simplify our answer further:

$t'=\frac{(2s^3+5s^2)-(2s^3+10s-2s)}{s^4}$

$t'=\frac{2s^3+5s^2-2s^3-10s+2s}{s^4}$

Thus, we combine our terms, do our last simplification, and reach our answer:

$t'=\frac{-5s^2+2s}{s^4}$

$t'=\frac{-5s+2}{s^3}$

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Example Question #451 : Derivatives

Find the derivative of the function:

$r=\frac{e^s}{s}$

Possible Answers:

$r' = -\frac{se^s+e^s}{s^2}$

$r' = \frac{s^se-e^2}{s^2}$

$r' = \frac{-se^s-e^s}{s}$

None of the other answers are correct

$r' = \frac{se^s-e^s}{s^2}$

Correct answer:

$r' = \frac{se^s-e^s}{s^2}$

Explanation:

To answer this question, we must understand two key concepts:

First, we must understand that the derivative of e^x is e^x , as long as the exponent is a variable by itself (no coefficients or other operations). Now, x can be replaced with any variable letter, really. So, the derivative of e^s is just e^s .

Second, we must understand the quotient rule of derivation. The quotient rule is very important to calculus and it looks like this:

$d/dx (\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g(x)}{g(x)^2}$

This rule shows you how to take the derivative of a quotient, hence the quotient rule.

(Here's a quick rhyme to remember this rule, if that helps:

"Low "dee" high less high "dee" low, ["dee" meaning derivative of]

draw the line and down below,

the denominator squared will surely go!" )

Combining the derivative of "e" raised to a variable, and the quotient rule, we can carry out our solution:

$r=\frac{e^s}{s}$

$r'=\frac{s(e^s)-e^s(1)}{s^2}$ (quotient rule)

$r'=\frac{se^s-e^s}{s^2}$

(you can leave the answer like this, or simplify further to the next line)

$r'=\frac{e^s(s-1)}{s^2}$

And you are done!

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Example Question #452 : Derivatives

Find the derivative of the function:

$h(x)=\frac{cos(x)}{sin^2(x)}$

Possible Answers:

$h'(x)=\frac{2}{{2sin^2(x)}}+\frac{cos^2(x)}{sin^2(x)}$

$h'(x)=\frac{-1}{{sin(x)}}+\frac{2cos^2(x)}{sin^3(x)}$

$h'(x)=\frac{1}{{sin^2(x)}}+\frac{2cos^2(x)}{sin^4(x)}$

$h'(x)=\frac{-1}{{sin(x)}}-\frac{2cos^2(x)}{sin^3(x)}$

$h'(x)=\frac{1}{{sin^2(x)}}-\frac{2cos^2(x)}{sin^4(x)}$

Correct answer:

$h'(x)=\frac{-1}{{sin(x)}}-\frac{2cos^2(x)}{sin^3(x)}$

Explanation:

*****************************************************************

REQUIRED KNOWLEDGE

***************************************************************** $1.\frac{d}{dx}(sin(x))=cos(x)$

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

$3. \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

$4. \frac{d}{dx}(f(x)*{g(x))}=f(x)g'(x)+g(x)f'(x)$

If you understand the first two derivatives, as well as number three and number four - the quotient and product rules for derivatives, respectively - this problem is not that difficult! However, it is easy to make a mistake keeping track of the term grouping (parentheses, brackets, and such). You can do it!

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SOLUTION STEPS

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Given:

$h(x)=\frac{cos(x)}{sin^2(x)}$

Set the problem up as a quotient rule:

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ (Required knowledge)

Thus:

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

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

We run into a small problem with the d(sin^2(x)) in the above line. To find the derivative of this term, we must use the product rule (or know the derivative from memory)

d(sin^2(x))=d(sin(x)*sin(x))

=sin(x)cos(x)+sin(x)cos(x)

=2sin(x)cos(x)

We can now continue the problem, plugging in the above portion for the d(sin^2(x))

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

$\frac{d}{dx}(\frac{cos(x)}{sin^2(x)})=\frac{[sin^2(x)(-sin(x))]-[cos(x)2sin(x)cos(x)]}{sin^4(x)}$

$\frac{d}{dx}(\frac{cos(x)}{sin^2(x)})=\frac{-sin^3(x)-cos(x)2sin(x)cos(x)}{sin^4(x)}$

$\frac{d}{dx}(\frac{cos(x)}{sin^2(x)})=\frac{-sin^3(x)-2cos^2(x)sin(x)}{sin^4(x)}$

$\frac{d}{dx}(\frac{cos(x)}{sin^2(x)})=\frac{-sin^3(x)}{{sin^4(x)}}-\frac{2cos^2(x)sin(x)}{sin^4(x)}$ break the terms up

Thus, our final, simplified answer should be:

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CORRECT ANSWER

*****************************************************************

$\frac{d}{dx}(\frac{cos(x)}{sin^2(x)})=h'(x)=\frac{-1}{{sin(x)}}-\frac{2cos^2(x)}{sin^3(x)}$

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PROBLEMS?

*****************************************************************

If you had problems with the concepts in this question, refer to the "required knowledge" at the top of the explanation for this problem. If you have any other questions on the algebra portion, or need clarification on anything done in this explanation, ask your Varsity Tutor for assistance or further explanation!

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Example Question #453 : Derivatives

$s = 9 + \frac{1}{cot(z)}$

Find the derivative of the function:

$s= 9 + \frac{1}{cot(z)}$

Possible Answers:

s'=-csc^2(z)

s'=-csc(z)cot(z)

s'=sec^2(z)

s'=-sec^2(z)

s'=-sec(z)

Correct answer:

s'=sec^2(z)

Explanation:

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REQUIRED KNOWLEDGE

*****************************************************************

$1. \frac{1}{cot(z)}=tan(z)$

$2. \frac{d}{dz}(tan(z))=sec^2(z)$

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SOLUTION STEPS

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Given:

$s= 9 + \frac{1}{cot(z)}$

$s'=\frac{d}{dz}( 9 + \frac{1}{cot(z)})$

$s'=\frac{d}{dz}( 9 )+\frac{d}{dz}( \frac{1}{cot(z)})$

$s'=0+\frac{d}{dz}(tan(z))$

$s'=\frac{d}{dz}(tan(z))$

Thus, our answer is:

*****************************************************************

CORRECT ANSWER

*****************************************************************

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

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Example Question #454 : Derivatives

Find the derivative of the function:

h(x)=(3+x)tan^2(x)

Possible Answers:

h'(x)=2tan(x)(6sec^2(x)-2xsec^2(x)-tan(x))

h'(x)=3sec^2(x)(6sec^2(x)+2xsec^2(x)+tan(x))

h'(x)=tan(x)(6sec^2(x)+2xsec^2(x)+tan(x))

h'(x)=-tan(x)(6sec^2(x)+2xsec^2(x)+tan(x))

$h'(x)=\frac{tan(x)(6sec^2(x)+2xsec^2(x)+tan(x))}{2}$

Correct answer:

h'(x)=tan(x)(6sec^2(x)+2xsec^2(x)+tan(x))

Explanation:

*****************************************************************

REQUIRED KNOWLEDGE

*****************************************************************

$1. \frac{d}{dx}(f(x)*g(x))=f(x)g'(x)+g(x)f'(x)$

$2. \frac{d}{dx}(tan(x))=sec^2(x)$

*****************************************************************

SOLUTION STEPS

*****************************************************************

Given:

h(x)=(3+x)tan^2(x)

Set up as product rule:

$h'(x)=[(3+x){\color{Blue} d(tan^2(x))]}+[tan^2(x)d(3+x)]$

Evaluate the derivative of tan^2(x) as another product rule, before continuing:

d(tan^2(x))=d(tan(x)*tan(x))

=tan(x)sec^2(x)+tan(x)sec^2(x)

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

Plug this into the slot in the initial product, and carry on

$h'(x)=[(3+x){\color{Blue} 2tan(x)sec^2(x)}]+[tan^2(x)d(3+x)]$

$h'(x)=[(3+x){\color{Blue} 2tan(x)sec^2(x)}]+[tan^2(x)(1)]$

h'(x)=6tan(x)sec^2(x)+2xtan(x)sec^2(x)+tan^2(x)

Pulling out the common factor of tan(x), we arrive at our answer:

*****************************************************************

CORRECT ANSWER

*****************************************************************

${\color{Red} h'(x)=tan(x)(6sec^2(x)+2xsec^2(x)+tan(x))}$

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Example Question #455 : Derivatives

Find the derivative of the function:

r(s)=s(3s^2-9)

Possible Answers:

r'(s)=-9(s^2-1)

r'(s)=(s^2+1)

r'(s)=9(s^2-1)

r'(s)=(s^2-1)

r'(s)=-(s^2-1)

Correct answer:

r'(s)=9(s^2-1)

Explanation:

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STEPS

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Given:

r(s)=s(3s^2-9)

Set the approach as a product rule application:

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

r'(s)=s(6s)+(3s^2-9)(1)

r'(s)=6s^2+3s^2-9

r'(s)=9s^2-9

Factoring out the 9, we then arrive at the correct answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} r'(s)=9(s^2-1)}$

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Example Question #456 : Derivatives

Find the derivative of the function:

y=(x-3)(3x+x^4)

Possible Answers:

y'=-(5x^4-12x^3)+6x-9

y'=5x^4+12x^3+6x+9

y'=5x^4+12x^3+6x-9

y'=5x^4-12x^3+6x-9

y'=5x^4+12x^3-6x+9

Correct answer:

y'=5x^4-12x^3+6x-9

Explanation:

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STEPS

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Given:

y=(x-3)(3x+x^4)

Identify the derivative as an application of the product rule

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

Continuing:

y'=[(x-3)(4x^3+3)]+[(3x+x^4)(1)]

y'=4x^4+4x-12x^3-9+3x+x^4

Combining like terms,

y'=5x^4+6x-12x^3-9

Rearranging the terms in descending degrees, we arrive at the correct answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} y'=5x^4-12x^3+6x-9}$

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Example Question #457 : Derivatives

Find the derivative of the function:

$f(x)=\frac{8x^2}{4-x}$

Possible Answers:

$f'(x)=-\frac{4x(4-x)}{(8-x)^2}$

$f'(x)=-\frac{x(8-x)}{(4-x)^3}$

$f'(x)=\frac{4x(4-x)}{(8-x)^2}$

$f'(x)=\frac{8x(8-x)}{(4-x)^2}$

$f'(x)=-\frac{x(4-x)}{(8-x)^3}$

Correct answer:

$f'(x)=\frac{8x(8-x)}{(4-x)^2}$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$f(x)=\frac{8x^2}{4-x}$

Set the derivative up with the quotient rule:

$f'(\frac{u}{v})=\frac{v(u')-u(v')}{v^2}$

$f'(x)=\frac{(4-x)(16x)-(8x^2)(-1)}{(4-x)^2}$

Multiply factors:

$f'(x)=\frac{64x-16x^2+8x^2}{(4-x)^2}$

Combine like terms:

$f'(x)=\frac{64x-8x^2}{(4-x)^2}$

Factoring out 8x in the numerator, we arrive at the correct answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} f'(x)=\frac{8x(8-x)}{(4-x)^2}}$

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Example Question #458 : Derivatives

Find the derivative of the function:

r(s)=s^3-s^2sin(s)

Simplify your final answer through factoring terms out

Possible Answers:

r'(s)=5s(s-cos(s)-2sin(s))

r'(s)=-5s(s+cos(s)+2sin(s))

r'(s)=s(3s-scos(s)-2sin(s))

r'(s)=3s(5s-3cos(s)-2sin(s))

$r'(s)=-\frac{1}{5s}(s+cos(s)+2sin(s))$

Correct answer:

r'(s)=s(3s-scos(s)-2sin(s))

Explanation:

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STEPS

*****************************************************************

Given:

r(s)=s^3-s^2sin(s)

r'(s)=d(s^3)-d(s^2sin(s))

r'(s)=3s^2-d(s^2sin(s))

Set up a product rule for the derivative of the second term:

$d(s^2sin(s))={\color{DarkBlue} [s^2(cos(s))+2s(sin(s))]}$

Plug this new-found product in for d(s^2sin(s)) ,

$r'(s)=3s^2- {\color{DarkBlue} d(s^2sin(s))}$

$r'(s)=3s^2-{\color{DarkBlue} [s^2(cos(s))+2s(sin(s))]}$

Multiply the negative across:

r'(s)=3s^2-s^2cos(s)-2s(sin(s))

Factor out a common factor of s, and we reach the correct answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} r'(s)=s(3s-scos(s)-2sin(s)){\color{Red} }}$

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Example Question #459 : Derivatives

Find the derivative of the function:

g(x)=3cos(x)-7sec(x)+5

Possible Answers:

$g'(x)=\frac{1}{3}[sin(x)+7sec(x)tan(x)]$

g'(x)=3sin(x)-7sec(x)tan(x)

g'(x)=3sec(x)+7sin(x)tan(x)

g'(x)=-3sin(x)-7sec(x)tan(x)

g'(x)=3sec(x)-7sin(x)tan(x)

Correct answer:

g'(x)=-3sin(x)-7sec(x)tan(x)

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

g(x)=3cos(x)-7sec(x)+5

We can avoid the product rule by factoring the constants out before derivation:

g'(x)=3d(cos(x))-7(d(sec(x)))+d(5)

g'(x)=(3)(-sin(x))-7d(sec(x))+0

g'(x)=-3sin(x))-7(d(sec(x))

Taking the derivative of the secant function, we arrive at the answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} g'(x)=-3sin(x)-7sec(x)tan(x)}$

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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 #71 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #451 : Derivatives

Example Question #452 : Derivatives

Example Question #453 : Derivatives

Example Question #454 : Derivatives

Example Question #455 : Derivatives

Example Question #456 : Derivatives

Example Question #457 : Derivatives

Example Question #458 : Derivatives

Example Question #459 : Derivatives

All AP Calculus AB Resources

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