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)#
