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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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200 Chapter 3 Differentiation Rules

ExamplE 2 Differentiate (a) y − sinsx 2 d and (b) y − sin 2 x.

SOLUTION

(a) If y − sinsx 2 d, then the outer function is the sine function and the inner function is

the squaring function, so the Chain Rule gives

dy

dx − d

dx

sin sx 2 d − cos sx 2 d ? 2x

outer

function

− 2x cossx 2 d

evaluated

at inner

function

derivative

of outer

function

evaluated

at inner

function

derivative

of inner

function

(b) Note that sin 2 x − ssin xd 2 . Here the outer function is the squaring function and the

inner function is the sine 7et0304note2

function. So

dy 01/13/10

dx − d

dx MasterID: ssin xd2 01593 − 2 ? ssin xd ? cos x

inner

function

inner derivative derivative evaluated

function of outer of outer at inner

function function

evaluatedderivative

at inner of inner

function function

derivative

of inner

function

See Reference Page 2 or Appendix D.

The answer can be left as 2 sin x cos x or written as sin 2x (by a trigonometric identity

known as the double-angle formula).

7et0304note3 7et0304note3

In Example 2(a) we combined 01/13/10 the 01/13/10 Chain Rule with the rule for differentiating the sine

function. In general, if y −MasterID: sin u, MasterID: where 01594 u is a differentiable 01594 function of x, then, by the

Chain Rule,

dy

dx − dy du du

− cos u

du dx dx

Thus

d

du

ssin ud − cos u

dx dx

In a similar fashion, all of the formulas for differentiating trigonometric functions can

be combined with the Chain Rule.

Let’s make explicit the special case of the Chain Rule where the outer function f is

a power function. If y − ftsxdg n , then we can write y − f sud − u n where u − tsxd. By

using the Chain Rule and then the Power Rule, we get

dy

dx − dy

du

du

dx

− nu

n21

du

dx − nftsxdgn21 t9sxd

4 The Power Rule Combined with the Chain Rule If n is any real number

and u − tsxd is differentiable, then

Alternatively,

d

dx su n n21

du

d − nu

dx

d

dx ftsxdgn − nftsxdg n21 ? t9sxd

Notice that the derivative in Example 1 could be calculated by taking n − 1 2 in Rule 4.

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