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.
 A: $f(x)=-6\sin^4 x=-6(\sin x)^4$
By Chain Rule,
$f'(x)=4\cdot-6(\sin x)^3(\sin x)'=-24(\sin x)^3\cos x=-24\sin^3x\cos x$
A: Steps:: Note this is a candidate for the chain rule. 
$$\text{Let } \space y = f(x) = -6\sin^4(x)$$
$$\text{Let } \space u=\sin(x) \Rightarrow y=-6u^4$$
To find $f'(x)$:
$$\frac{\mathrm{d}y}{\mathrm dx}=\frac{\mathrm dy}{\mathrm du}\cdot\frac{\mathrm du}{\mathrm dx}\tag{1}$$
$$\frac{\mathrm dy}{\mathrm dx}=-6(4)u^3\cdot \cos(x)= -24\sin^3(x)\cos(x)$$
$(1)$ can also be expressed as $(f \circ g)'(x)=f'(g(x))\cdot g'(x)$ Where $g(x)$ is the inner function $\sin(x)$ and $f(u)$ is the outer function $-6u^4$
A: Whether $f(x) = -6\sin^4 x = -6(\sin x)^4$ or $f(x) = -6 \sin (x^4)$, the chain rule is the way to go.
If
$f(x) = -6(\sin x)^4, \tag{1}$
set
$u = \sin x \tag{2}$
so that (1) becomes
$f(u(x)) = -6(u(x))^4; \tag{3}$
now we apply the chain rule, which says that $df(u(x))/dx$ is given by
$\dfrac{df(u(x))}{dx} = \dfrac{df}{du}(u(x)) \dfrac{du}{dx}(x). \tag{4}$
We see that
$\dfrac{df(u(x))}{du}(u(x)) = -24(u(x))^3 = -24 (\sin x)^3 \tag{5}$
and
$\dfrac{du}{dx} = \cos x, \tag{6}$
whence, by (4),
$\dfrac{df(u(x))}{dx} = -24 (\sin x)^3 \cos x. \tag{7}$
On the other hand if
$f(x) = -6\sin(x^4), \tag{8}$
then we take
$u(x) = x^4 \tag{9}$
so that (8) yields
$f(u(x)) = -6 \sin u(x), \tag{10}$
from which we have
$\dfrac{df(u(x))}{du}(u(x)) = -6 \cos u(x) = -6 \cos x^4; \tag{11}$
also,
$\dfrac{du}{dx} = 4x^3; \tag{12}$
pulling (11) and (12) together with the aid of (4) gives the result
$\dfrac{df((x))}{dx} = -24x^3 \cos x^4. \tag{13}$
Either way, The Chain Rule rules!  Long Live The Chain Rule! 
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: 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 $g(x) = -6 \sin x^4$:
$$f'(x) = -6 \frac{d}{dx} \left(\sin^4 x\right) = -24 \sin^3x \frac{d}{dx} \sin x = -24 \sin^3x \cos x.$$
$$g'(x) = -6 \frac{d}{dx} \left(\sin x^4\right) = - 6 \cos x^4 \frac{d}{dx} x^4 = -24 x^3 \cos x^4.$$
