Question

How do you derive `f(x)=x^5*3^(x^2-3*x+2)`?

Answer 1

`f'(x) = (5+(2x^2-3x)ln3)( x^4*3^(x^2-3x+2) )`

Explanation:

We have:

`f(x) = x^5*3^(x^2-3x+2)`

Take Natural Logarithms of both sides:

`ln f = ln {x^5*3^(x^2-3x+2)}`
`" " = ln (x^5) + ln (3^(x^2-3x+2))`
`" " = (5ln x) + (x^2-3x+2)ln3`

Differentiating Implicitly, we get:

`1/f * (df)/(dx) = 5/x+(2x-3)ln3`
`" " = (5+(2x-3)xln3)/x`
`" " = (5+(2x^2-3x)ln3)/x`

And so:

`(df)/(dx) = (5+(2x^2-3x)ln3)/xf`
`" = " (5+(2x^2-3x)ln3)/x x^5*3^(x^2-3x+2)`
`" = " (5+(2x^2-3x)ln3)( x^4*3^(x^2-3x+2) )`