Question

g(x)=e^x/(1-4x) find g'(x)

ive gotten to ((1-4x)(e^x)-(e^x)(-4))/(1-4x)^2 but i'm having difficulty further simplifying this

Answer 1

What's preventing you from expanding the numerator and hoping there might be some simplification?

Answer 2

the exponentials are confusing me or else i would already expanded and simplified it

Answer 3

Actually, I might be tempted to observe: \(\dfrac{(1-4x)\cdot e^{x} + 4\cdot e^{x}}{(1-4x)^{2}} = \dfrac{e^{x}}{1-4x}\cdot\dfrac{(1-4x) + 4}{1-4x}\)

This leads to the rather interesting: \(g'(x) = g(x)\cdot\dfrac{5-4x}{1-4x}\).

It's almost lovely!