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)=

Question

asked Mar 4, 2021 187k views

1 Answer

Stephen Mulcahy · Answer 1 · 2021-03-08T17:38:30+0000

Answer:

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

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:

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