Answer:
[tex]h'(e) = 7^{e-1}\cdot [7\cdot \ln 7+6\cdot e][/tex]
Step-by-step explanation:
Let [tex]h(x) = f(x)^{x}[/tex], the first derivative of the function is found by applying the concept of implicit differentiation:
[tex]h(x) = f(x)^{x}[/tex] (1)
[tex]\ln h(x) = x\cdot \ln f(x)[/tex]
[tex]\frac{h'(x)}{h(x)}=\ln f(x) +\frac{x\cdot f'(x)}{f(x)}[/tex]
[tex]h'(x) = h(x) \cdot \left[\ln f(x)+\frac{x\cdot f'(x)}{f(x)} \right][/tex]
[tex]h'(x) = f(x)^{x}\cdot \left[\ln f(x)+\frac{x\cdot f'(x)}{f(x)} \right][/tex]
[tex]h'(x) = f(x)^{x-1}\cdot [f(x)\cdot \ln f(x)+x\cdot f'(x)][/tex] (2)
If we know that [tex]x = e[/tex], [tex]f(e) = 7[/tex] and [tex]f'(e) = 6[/tex], then [tex]h'(e)[/tex] is:
[tex]h'(e) = 7^{e-1}\cdot [7\cdot \ln 7+6\cdot e][/tex]
