Question

Find y' for y=(x)^(x^2) (derivative)

Answer 1

`y=x^(x^2)`

`\ln y=\ln x^(x^2)`

`\ln y=x^2\ln x`

Now differentiating with respect to
`x` gives


`\frac{y'}y=2x\ln x+\frac{x^2}x`

`\frac{y'}y=2x\ln x+x`

`\frac{y'}y=x(\ln x^2+1)`

`y'=yx(\ln x^2+1)`

`y'=x^(x^2)x(\ln x^2+1)`

`y'=x^(x^2+1)(\ln x^2+1)`