The derivative of f (x) can be calculated using chain rule that says:
f (x) can be written as composite functions where:
v (x)=e^(2x)-3lnx
u (x)=x^4
So,
f (x)=u (v (x))
Applying chain rule on the composite function f (x)we have:
color (purple)(f'(x)=u (v (x))'
color (purple)(f'(x)=v'(x)×u'(v (x)))
Let's find color (purple)(v'(x)
Applying chain rule on the derivative of exponential:
color (red)((e^(g (x)))'=g'(x)×e^(g (x)))
Knowing the derivative of ln (x) that says:
color (brown)((ln (g (x)))'=(g'(x))/(g(x)))
color (purple)(v'(x))=color (red)((2x)'e^(2x))-3color (brown)((x')/(x))
color (purple)((v'(x))=2e^(2x)-(3/x))
Let's find color (blue)(u'(x)):
Applying the derivative of power stated as follows:
color (green)(x^n=nx^(n-1)
color(blue)(u'(x))=color (green)(4x^3)
Based on chain rule above we need u'(v (x)) so let's substitute x by v (x):
u'(v (x))=4 (v (x))^3
color (purple)(u'(v (x))=4 (e^(2x)-3lnx)^3)
Let's substitute the values of u'(v (x))and v'(x) in the above chain rule above we have:
color (purple)(f'(x)=v'(x)×u'(v (x)))
color (purple)(f'(x)=(2e^(2x)-(3/x))× 4 (e^(2x)-3lnx)^3)
color (purple)(f'(x)=4 (2e^(2x)-(3/x))× (e^(2x)-3lnx)^3)