Question

find f'x for f(x)= (3x-sin^2(4x))^1/2

Answer 1

Alright...you have:
\[f(x)=(3x-\sin^2(4x))^{\frac{1}{2}}=\sqrt{3x-\sin^2(4x)}\]
You are going to be using the chain rule a bit here; i'll simplify a little if you don't mind:
\[\eqalign{
f(x)&=\sqrt{3x-\sin^2(4x)} \\
&=\sqrt{3x-[\sin(4x)]^2} \\
&=\sqrt{3x-[g(x)]^2} \\
&=\sqrt{3x-h(x)} \\
&=\sqrt{j(x)} \\
}\]
So then we can expand a little:
\[\eqalign{
f'(x)&=\frac{d}{dx}\sqrt{j(x)} \\
&=\frac{1}{2\sqrt{j(x)}}\times j'(x) \\
&=\frac{1}{2\sqrt{j(x)}}\times[3-h'(x)] \\
&=\frac{1}{2\sqrt{j(x)}}\times[3-2g(x)\times g'(x)] \\
&=\frac{1}{2\sqrt{j(x)}}\times[3-2g(x)\times 4\sin(x)] \\
&=\frac{3-2g(x)\times 4\sin(x)}{2\sqrt{j(x)}} \\
&=\frac{3-8\sin(4x)\sin(x)}{2\sqrt{3x-\sin^2(4x)}}
}\]

Answer 2

hmmm Im a bit confused by the first part