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aleksandrvk [35]
3 years ago
8

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

Mathematics
1 answer:
arlik [135]3 years ago
6 0

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]

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