Chapter 3 | Derivatives
255
To check, we see that j ( x ) =3 x 5 + x 3 −10 x and, consequently, j ′( x ) =15 x 4 +3 x 2 −10.
Use the product rule to obtain the derivative of j ( x ) =2 x 5 ⎛
⎝ 4 x 2 + x ⎞ ⎠ .
3.16
The Quotient Rule Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that d dx ⎛ ⎝ x 2 ⎞ ⎠ =2 x , not d dx ⎛ ⎝ x 3 ⎞ ⎠ = 3 x 2 1 =3 x 2 .
d dx ( x )
Theorem 3.6: The Quotient Rule Let f ( x ) and g ( x ) be differentiable functions. Then
⎛ ⎝ f ( x ) g ( x )
⎞ ⎠ =
d dx ( f ( x )) · g ( x )−
d dx ( g ( x )) · f ( x )
d dx
.
( g ( x )) 2
That is,
f ′( x ) g ( x )− g ′( x ) f ( x ) ( g ( x )) 2 .
f ( x ) g ( x ) , then
if j ( x ) =
j ′( x ) =
The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example. Example 3.25 Applying the Quotient Rule
Use the quotient rule to find the derivative of k ( x ) = 5 x 2 4 x +3 .
Solution Let f ( x ) =5 x 2 and g ( x ) =4 x +3. Thus, f ′( x ) =10 x and g ′( x ) =4. Substituting into the quotient rule, we have k ′( x ) = f ′( x ) g ( x )− g ′( x ) f ( x ) ( g ( x )) 2 = 10 x (4 x +3)−4(5 x 2 ) (4 x +3) 2 .
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