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

Answer 1

Answer:

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

Step-by-step 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)

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

$\frac{h'(x)}{h(x)}=\ln f(x) +\frac{x\cdot f'(x)}{f(x)}$

$h'(x) = h(x) \cdot \left[\ln f(x)+\frac{x\cdot f'(x)}{f(x)} \right]$

$h'(x) = f(x)^{x}\cdot \left[\ln f(x)+\frac{x\cdot f'(x)}{f(x)} \right]$

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

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

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