neato
write √sin(8x) as [tex](sin(8x))^\frac{1}{2}[/tex]
then use the power rule
remember
[tex]\frac{dy}{dx}x^m=mx^{m-1}[/tex]
and the chain rule
[tex]\frac{dy}{dx}f(g(x))=f'(g(x))g'(x)[/tex]
and derivitive of sin(x)=cos(x)
so
[tex]\frac{dy}{dx} \sqrt{sin(8x)} [/tex]=
[tex]\frac{dy}{dx} (sin(8x))^\frac{1}{2} [/tex]=
[tex]\frac{1}{2}(sin(8x))^\frac{-1}{2}\frac{dy}{dx}sin(8x) [/tex]=
[tex]\frac{1}{2\sqrt{sin(8x)}}cos(8x)\frac{dy}{dx}8x [/tex]=
[tex]\frac{8cos(8x)}{2\sqrt{sin(8x)}}[/tex]=
[tex]\frac{4cos(8x)}{\sqrt{sin(8x)}}[/tex]=
answer is 2nd option
