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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

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REQUIRED KNOWLEDGE

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

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

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SOLUTION STEPS

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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:

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CORRECT ANSWER

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

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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:

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ANSWER

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

Report an Error

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:

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ANSWER

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

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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:

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STEPS

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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:

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ANSWER

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

Report an Error

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

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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:

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ANSWER

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

Report an Error

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:

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STEPS

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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:

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ANSWER

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

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

Report an Error

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

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

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

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

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

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

Report an Error

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

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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

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

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

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

Report an Error

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

Find the derivative of the function:

x=csc(y)cot(y)

Possible Answers:

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

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

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

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

x'=2csc(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

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

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

Report an Error

Example Question #84 : 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)=sin(r)sin(r)

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

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

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:

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

ANSWER

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

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

Report an Error

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #454 : Derivatives

Example Question #455 : Derivatives

Example Question #456 : Derivatives

Example Question #457 : Derivatives

Example Question #458 : Derivatives

Example Question #459 : Derivatives

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

Example Question #84 : Derivative Rules For Sums, Products, And Quotients Of Functions

All AP Calculus AB Resources

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