CK-12-Calculus

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 2. Derivatives


f′(x) =x^2 (−sinx)+ 2 xcosx+cosx
=−x^2 sinx+ 2 xcosx+cosx.

Example 3:
Finddy/dxify= 1 −costanxx. What is the slope of the tangent line atx=π/3?
Solution:
Using the quotient rule and the formulas above, we obtain


dy
dx=

( 1 −tanx)(−sinx)−(cosx)(−sec^2 x)
( 1 −tanx)^2
=−sinx+tanxsinx+cosxsec

(^2) x
( 1 −tanx)^2
=−sinx+( 1 tan−tanxsinx)x 2 +secx
To calculate the slope of the tangent line, we simply substitutex=π/3:
dy
dx


∣∣


∣∣


∣x=π 3 =

−sin(π 3 )+tan(π 3 )sin(π 3 )+sec(π 3 )
( 1 −tan(π 3 ))^2.

We finally get the slope to be approximately


dy
dx

∣∣


∣∣


∣x=π 3 =^4.^9.

Example 4:
Ify=secx, findy′′(π/ 3 ).
Solution:


y′=secxtanx
y′′=secx(sec^2 x)+(secxtanx)tanx
=sec^3 x+secxtan^2 x.

Substituting forx=π/3,


y′′=sec^3


3

)


+sec


3

)


tan^2


3

)


= ( 2 )^3 +( 2 )(



3 )^2


= 8 +( 2 )( 3 )


= 14.


Thusy′′(π/ 3 ) =14.

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