$$\text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. }$$
-$$f ( u ) = \sin ^ { 2 } \pi u + u , u = g ( x ) = - x , x = 6$$
A) 1
B) 0
C) -6
D) -1

Question

Quizplus · Accepted Answer

$(f \circ g) (x) = f(g(x)) = f(-x) = \sin^2(-\pi x) - x$

$(f \circ g)'(x) = \frac{d}{dx}[\sin^2(-\pi x) - x]$

Using the chain rule and derivative of sine function, we get:

$(f \circ g)'(x) = 2\sin(\pi x)\cdot\pi\cdot(-1)-1 = -2\pi\sin(\pi x)-1$

At $x=6$, we have:

$(f \circ g)'(6) = -2\pi\sin(6\pi)-1 = -2\pi\cdot 0 -1 = -1$

Find the value of (f∘g)′ at the given value of x. \text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. } Find the value of (f∘g)′ at the given value of x.

$\text { Find the value of } ( f \circ g ) ^ { \prime } \text { at the given value of } x \text {. }$