Question

find f'(x)
y=tan^4(x^3)

i got (4sec^2(x^3))^3*3x^2 am i right

Answer 1

\(\large { \bf
\cfrac{d}{dx}[tan(x^3)]^4\implies 4[tan(x^3)]^3\cdot [sec(x^3)]^2\cdot 3x^2
\\ \quad \\
\implies 4tan^3(x^3)sec^2(x^3)3x^2
}\)