Question

Please help
f(x) = -8 cos ( sin ( x ^{4} ))
find f'(x)

Answer 1

Okay so im just going to identify composite functions:
\[f(x)=-8\cos[\sin(x^4)]=-8\cos[\sin(g(x))]=-8\cos[h(x)]=-8j(x)\]
So then we have simplified the question to:
\[\eqalign{
\frac{d}{dx}f(x)&=[-8j(x)]' \\
&=-8j'(x) \\
&=-8[\cos(h(x))]' \\
&=-8[-\sin(h(x))h'(x)] \\
&=8\sin[h(x)]h'(x) \\
&=8\sin[\sin(g(x))][\sin(g(x))]' \\
&=8\sin[\sin(x^4)][\cos(g(x))g'(x)] \\
&=8\sin[\sin(x^4)]\cos[g(x)]g'(x) \\
&=8\sin[\sin(x^4)]\cos(x^4)(x^4)' \\
&=8\sin[\sin(x^4)]\cos(x^4)(4x^3) \\
&=32x^3\sin[\sin(x^4)]\cos(x^4) \\
}\]

Answer 2

Thanks, that was helpful

Answer 3

No prob!