Question

How do you find the derivative of `y=sqrt(3^x-2^x)`?

Answer 1

`=1/(2y)(3^xln3-2^xln2 )`

Explanation:

Use `a^x=e^(lna x)`

`y=(e^(ln3 x)-e^(ln2 x))^(1/2)`

`y'=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(e^(ln3 x)-e^(ln2 x))'`

`=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(ln3e^(ln3 x)-ln2e^(ln2 x))`

`=1/(2y)(3^xln3-2^xln2 )`