MATH241 Lecture Notes - Lecture 15: Quotient Rule, Product Rule
9 May 2018
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MATH241 - Lecture 15 - Derivatives of Trigonometric Functions and The Chain Rule
3.3: Derivatives of Trigonometric Functions (conclusion)
Example
:
Find the equation of the tangent line of at (x)cosxf =x+ 2 x=2
π
Solution:
Slope: (x)sinxf′= 1 − 2
sin m=f′(2
π)= 1 − 2 2
π
(1)= 1 − 2
= − 1
x1=2
π
(x) y1=f1=f(2
π)cos=2
π+ 2 2
π=2
π
Point-Slope: (x)y−y1=m−x1
− y−2
π= 1 x
(−2
π)
−y−2
π=x+2
π
−y=x+ π
Multiple Product/Quotient Rules
Let F=f·g·h
fgh] (fg)h]F′= [ ′= [ ′
f g)h fg)h= ( ′+ ( ′
f g )h= ( ′+fg′+fgh′
gh h=f′+fg′+fgh′
Example
:
Differentiate (x)e cotxf =x2x
Solution:
(x)xe cotx e cotx e (− sc x)f′= 2 x+x2x+x2xc2
=xex(2 cotx cotx csc x)+ x−x2
Example
:
Differentiate (x)g=xex
sinx
Solution:
(x)g′=sin x
2
(xe ) sinx − xe cosx
x′x
=sin x
2
(1)e + xe sinx − xe cosx[x x]x
=sin x
2
e sinx + xe sinx − xe cosx
x x x
=sin x
2
e(sinx + x sinx − x cosx)
x
3.4: The Chain Rule
We can find the derivative of by expanding (x)xf = ( 2+ 1)2(x)f
(x)xf = ( 2+ 1)2
x=x4+ 2 + 1
(x)xf′= 4x3+ 4
x (x)= 4 2+ 1
To find the derivative of it is not necessary to expand . Notice that is a(x)xf = ( 2+ 1)2(x)f(x)f
composition of two functions: and , which we can differentiate with the(x)g=x2(x)h=x2+ 1
rules we already know.
The Chain Rule allows us to determine the derivatives of long expressions such as
, functions that cannot be simplified such as , and even compositex(3+x2+ 1)20 x)( 3+xe−x2
3
functions like .ins (cos (x))
2 3 + 1
The Chain Rule
If and are both differentiable and , then the derivative of is thef g (g(x))F=f·g=f F
product of the derivative of the “outer” function and the “inner” function f g
(g(x))g(x)F′=f′′
Example
:
Differentiate
A) (x)xf = ( 2+ 1)2
