Dirty differentiation

First of all, the problem should be edited to d/dt because we have to take the derivative of the equation with respect to t and not x.

d/dt (cos^4(t)) + d/dt(cos(t)^4)

Using product rule and chain rule respectively,

we get

(-4 sin t )(cos^3 t ) - (4t^3)(sin(t^4)) [Ask me in the comments section if you have any problem getting it]

Therefore, -4 + 3 + 4 + 4 = 7

rather than just solving the problem , we try to take the generalisation $\frac{d}{dx}\cos^a(bx^c)$ We can generalise the chain rule for 3 or more functions (here there are 4 functions)which we can get by repeated application of chain rule . $(f\circ g\circ h\circ i)'(x)=f'(( g\circ h\circ i)(x))\cdot (g\circ h\circ i)'(x)$ $=f'((g\circ h\circ i)(x))\cdot g'((h\circ i)(x))\cdot (h\circ i)'(x)$ $=f'((g\circ h\circ i)(x))\cdot g'((h\circ i)(x))\cdot h'(i(x))\cdot i'(x)$ .This we can apply in above generalisation to obtain $\frac{d}{dx}\cos^a(bx^c)=-abcx^{c-1}\sin(bx^c)\cos^{a-1}(bx^c)$ which gives our answer as $(-4\sin(t)\cos^3(t))-(4t^3)(\sin(t^4))$ .

If you have any queries you can discuss in comment box .