Calculus- Early Transcendentals, 2021a

Now we need the derivative of
So the original derivative is
√
d
1 +
dx
√
1 + √ x. Using the chain rule again:
√
d
1 + √ x = 1 ( √ ) −1/2 1
1 + x
dx
2
2 x−1/2 .
√
1 + √ x = 1 ( √
1 + 1 + √ ) −1/2
1
x
2
2
1
=
8 √ x √ 1 + √ √
x 1 + √ 1 + √ x
(
1 +
√ x
) −1/2 1
2 x−1/2 .
4.5. The Chain Rule 139
Using the chain rule, the power rule, and the product rule, it is possible to avoid using the quotient rule
entirely.
Example 4.34: Derivative of Quotient without Quotient Rule
Compute the derivative of f (x)=
x3
x 2 + 1 .
♣
Solution. Write f (x)=x 3 (x 2 + 1) −1 ,then
f ′ (x) = x 3 d dx (x2 + 1) −1 + 3x 2 (x 2 + 1) −1
= x 3 (−1)(x 2 + 1) −2 (2x)+3x 2 (x 2 + 1) −1
= −2x 4 (x 2 + 1) −2 + 3x 2 (x 2 + 1) −1
=
=
−2x 4
(x 2 + 1) 2 + 3x2
x 2 + 1
−2x 4
(x 2 + 1) 2 + 3x2 (x 2 + 1)
(x 2 + 1) 2
= −2x4 + 3x 4 + 3x 2
(x 2 + 1) 2 = x4 + 3x 2
(x 2 + 1) 2
Note that we already had the derivative on the second line; all the rest is simplification. It is easier to get
to this answer by using the quotient rule, so there’s a trade off: more work for fewer memorized formulas.
♣