Question

Differentiate (a) y = sin(x^9) and (b) y = (sin(x))^9. SOLUTION (a) If y = sin(x^9), then the outer function is the sine function and the inner function is the power function, so the Chain Rule gives dy/dx = d/dx sin(x^9) = cos(x^9) * (b) Note that (sin(x))^9 = (sin (x)). Here the outer function is the power function and the inner function is the sine function. So dy/dx = d/dx (sin(x))^9 = * cos(x)

          Differentiate (a) y = sin(x^9) and (b) y = (sin(x))^9.

SOLUTION (a) If y = sin(x^9), then the outer function is the sine function and the inner function is the power function, so the Chain Rule gives

dy/dx = d/dx sin(x^9) = cos(x^9) *

(b) Note that (sin(x))^9 = (sin (x)). Here the outer function is the power function and the inner function is the sine function. So

dy/dx = d/dx (sin(x))^9 = * cos(x)

Differentiate (a) y = sin(x^9) and (b) y = (sin(x))^9.

SOLUTION (a) If y = sin(x^9), then the outer function is the sine function and the inner function is the power function, so the Chain Rule gives

dy/dx = d/dx sin(x^9) = cos(x^9) *

(b) Note that (sin(x))^9 = (sin (x)). Here the outer function is the power function and the inner function is the sine function. So

dy/dx = d/dx (sin(x))^9 = * cos(x)

Added by Jennifer O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ben Blakesley

09/30/2022

Step 1

Let's make the substitution \(u = x^9\), so \(y = \sin(u)\). ** Show more…

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Differentiate (a) y = sin(x^9) and (b) y = (sin(x))^9. SOLUTION (a) If y = sin(x^9), then the outer function is the sine function and the inner function is the power function, so the Chain Rule gives dy/dx = d/dx sin(x^9) = cos(x^9) * (b) Note that (sin(x))^9 = (sin (x)). Here the outer function is the power function and the inner function is the sine function. So dy/dx = d/dx (sin(x))^9 = * cos(x)