Derivative of $f(x)=-6\sin^4 x$ 
$f(x)=-6\sin^4x$  

$f(x)=-6\sin x^4$  
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$  
What am I doing wrong? Please show the steps.

As others have mentioned, $\sin^4x$ and $\sin x^4$ are completely different.
By convention, $\sin^4x = (\sin x)^4$, the fourth power of $\sin x$.
Also by convention, $\sin x^4 = \sin (x^4)$, the sine of the fourth power of $x$.
But here are derivatives for both $f(x) = -6 \sin^4 x$ and $g(x) = -6 \sin x^4$:
$$f'(x) = -6 \frac{d}{dx} \left(\sin^4 x\right) = -24 \sin^3x \frac{d}{dx} \sin x = -24 \sin^3x \cos x.$$
$$g'(x) = -6 \frac{d}{dx} \left(\sin x^4\right) = - 6 \cos x^4 \frac{d}{dx} x^4 = -24 x^3 \cos x^4.$$
