How to differentiate $y=\frac{2^x+4^x}{3^x+5^x}$

Question

Differentiate

$$y=\dfrac{2^x+4^x}{3^x+5^x}$$

I think you have to use implicit differentiation, but I don't know how to start. I first ln both sides and separated the fraction into subtraction because of the ln property.

Answer 1

$$\frac{d}{dx}\left[a^x\right]=a^x\ln a$$

$$\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u'v-v'u}{v^2}$$

Now put $u=2^x+4^x$ , $v=3^x+5^x$ $$y= \frac{2^x+4^x}{3^x+5^x}$$

$$\implies y'=\frac{(2^x\ln2+4^x\ln4)(3^x+5^x)-(3^x\ln3+5^x\ln5)(2^x+4^x)}{(3^x+5^x)^2}$$

Answer 2

$$y=\dfrac{2^x+4^x}{3^x+5^x}$$

$$y'=\dfrac{(3^x+5^x)(2^x ln(2)+4^xln(4))+(2^x+4^x)(3^x ln(3)+5^xln(5))}{(3^x+5^x)^2}$$

Simple Quotient rule application.

Using the fact that $$(\dfrac{f}{g})'=\dfrac{f'g+g'f}{g^2}$$

and

$$(n^x)'=n^x*ln(n)$$