Question

Consider the function f(x) for which f(e)=7 and f'(e)=6 find h'(e) for the function h(x)=f(x)^x

h'(e)=

Answer 1

`h'(e) = 7^(e-1)\cdot [7\cdot \ln 7+6\cdot e]`

Explanation:

Let
`h(x) = f(x)^(x)`, the first derivative of the function is found by applying the concept of implicit differentiation:

`h(x) = f(x)^(x)` (1)

`\ln h(x) = x\cdot \ln f(x)`

`(h'(x))/(h(x))=\ln f(x) +(x\cdot f'(x))/(f(x))`

`h'(x) = h(x) \cdot \left[\ln f(x)+(x\cdot f'(x))/(f(x)) \right]`

`h'(x) = f(x)^(x)\cdot \left[\ln f(x)+(x\cdot f'(x))/(f(x)) \right]`

`h'(x) = f(x)^(x-1)\cdot [f(x)\cdot \ln f(x)+x\cdot f'(x)]` (2)

If we know that
`x = e`,
`f(e) = 7` and
`f'(e) = 6`, then
`h'(e)` is:

`h'(e) = 7^(e-1)\cdot [7\cdot \ln 7+6\cdot e]`