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8

MORE RULES

FOR

DERIVATIVES

The quotient rule

The quotient rule

The following is called the quotient rule:

"The derivative of the quotient of two functions is equal to

the denominator times the derivative of the numerator
minus the numerator times the derivative of the denominator
all divided by the square of the denominator."

For example, accepting for the moment that the derivative of sin x is cos x (Lesson 12):

   Problem 1.   Calculate the derivative of      x2  
sin x
.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

sin x· 2xx2 cos x
         sin2x
   Problem 2.   Use the chain rule to calculate the derivative of   sin2x
  x3
.
x³· 2 sin x cos x − sin2x· 3x2
                 x6
  =   x2 sin x(2x cos x − 3 sin x)
                x6
 
    =   sin x(2x cos x − 3 sin x)
               x4
   Problem 3.   Calculate the derivative of    x2 − 5x − 6
    2x + 1
.
(2x + 1)(2x − 5) − (x2 − 5x − 6)· 2
                  (2x + 1)2
 =  4x2 − 8x − 5 − 2x2 + 10x + 12
                  (2x + 1)2
 
   =  2x2 + 2x + 7
   (2x + 1)2
   Problem 4.   Calculate the derivative of    3x2x + 4
     
.

See the Example, Lesson 6, and Lesson 22 of Algebra.

 = 
 
   = 
 
   = 

Proof of the quotient rule

THEOREM.

Proof.   Since g = g(x), then

 d 
dx 
1
g
 =    d 
dg
  1
g
·   dg 
dx
 =  − 1
g2
 g'

according to the chain rule, and Problem 4 of Lesson 5.

Therefore, according to the product rule (Lesson 6),

This is the quotient rule, which we wanted to prove.


Implicit differentiation

Consider the following:

x2 + y2 = r2

This is the equation of a circle with radius r. (Lesson 17 of Precalculus.)

  Let us calculate    dy
dx
.   To do that, we could solve for y and then take the

derivative.  But rather than do that, we will take the derivative of each term.  As for y2, we consider it implicitly a function of x, and therefore

  we may apply the chain rule to it.  Then we will solve for  dy
dx
.
 d 
dx
  x2   +    d 
dx
  y2   =    d
dx
  r2
2x + 2y   dy 
dx
  =   0
 
dy 
dx
  =   x
y
.

This is called implicit differentiation.  We treat y as a function of x and apply the chain rule.  The derivative that results generally contains both x and y.

Problem 5.   15y + 5y3 + 3y5 = 5x3.  Calculate y'.

15y' + 15y2y' + 15y4y' = 15x2
 
y'(1 + y2 + y4) = x2
y' =         x2      
1 + y2 + y4
   Problem 6.      Calculate y'.
= 0
 
=
 
y' =

Problem 7.

a)  In this circle,

x2 + y2 = 25,

a)   what is the y-coördinate when x = −3?

y = 4 or −4. For,

(−3)2 + (±4)2 = 52

b)  What is the slope of the tangent to the circle at (−3, 4)?

3
4
.  For, the derivative is − x
y
.

c)  What is the slope of the tangent to the circle at (−3, −4)?

3
4

Problem 8.   In the first quadrant, what is the slope of the tangent to this circle,

(x − 1)2 + (y + 2)2 = 169,

when x = 6?

[Hint:  52 + 122 = 132 is a Pythagorean triple.]

In the first quadrant, when x = 6, y = 10.

(6 − 1)2 + (10 + 2)2 = 132.

y' = − x − 1
y + 2
.  Therefore the slope is −   6 − 1
10 + 2
  =  −  5 
12

Problem 9.   Calculate the slope of the tangent to this curve at (2, −1):

x3 − 3xy2 + y3 = 1

  3x2 − (3x· 2y y' + y2· 3) + 3y2 y'  =  0
 
according to the product rule.
 
  3x2 − 6xy y' − 3y2 + 3y2 y'  =  0
 
  x2 − 2xy y'y2 + y2 y'  =  0
 
  y'(y2 − 2xy)  =  y2x2
 
  y'  =   y2x2 
y2 − 2xy
 
 Therefore, at (2, −1):
 
  y'  =      (−1)2 − 22    
(−1)2 − 2· 2· −1
 
     =  −3
  5
 
     =  3
5

The derivative of an inverse function

When we have a function  y = f(x) -- for example

y = x2

-- then we can often solve for x.  In this case,

On exchanging the variables, we have

is called the inverse function of  yx2.

Let us write

f(x) = x2
 
g(x) =

And let us call f the direct function and g the inverse function.  The formal relationship between f and g is the following:

f( g(x)) = g( f(x)) = x.

(Topic 19 of Precalculus.)

Here are other pairs of direct and inverse functions:

f(x) = sin x   g(x) = arcsin x
 
f(x) = ax   g(x) = logax
 
f(x) = x3   g(x) =

Now, when we know the derivative of the direct function f, then from it we can determine the derivative of g.

Thus, let g(x) be the inverse of f(x).  Then

f(g(x)) = x.

Now take the derivative with respect to x:

This implies the following:

Theorem.   If g(x) is the inverse of f(x), then

"The derivative of an inverse function is equal to

the reciprocal of the derivative of the direct function

when its argument is the inverse function."

   Example.   Let f(x) = x2,  and    Then  f( g) = g2.  

Therefore,

Next Lesson:  Instantaneous velocity and rates of change


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