1. ## Logarithmic differentiation

Logarithmic differentiation
Having some problems with these question....

f(x)
and g(x) are two differentiable functions and f(x)>0 for all x.
Consider h(x) = f(x)^g(x)

a) Show that ln h(x) = g(x) ln f(x)

b) Show that the derivate of ln h(x) with respect to x is:
g´(x) ln f(x) + g(x) f´(x)/f(x)

c) Show that the derivate of ln h(x) with respect to x is:
h´(x)/h(x)

d) Show that h´(x) = f(x)^g(x) [g´(x) ln f(x) + g(x) f´(x)/f(x)]

e) Use the above formula to find the derivate of h(x)=x^x.
Identify the functions f(x) and g(x) and their derivatives.

Thanks for help!! This is killing me...

2. Originally Posted by Lippi86
Logarithmic differentiation
Having some problems with these question....

f(x)
and g(x) are two differentiable functions and f(x)>0 for all x.
Consider h(x) = f(x)^g(x)

a) Show that ln h(x) = g(x) ln f(x)

b) Show that the derivate of ln h(x) with respect to x is:
g´(x) ln f(x) + g(x) f´(x)/f(x)

c) Show that the derivate of ln h(x) with respect to x is:
h´(x)/h(x)

d) Show that h´(x) = f(x)^g(x) [g´(x) ln f(x) + g(x) f´(x)/f(x)]

e) Use the above formula to find the derivate of h(x)=x^x.
Identify the functions f(x) and g(x) and their derivatives.

Thanks for help!! This is killing me...
a) $h(x) = f(x)^{g(x)}$

$\ln{\left[h(x)\right]} = \ln{\left[f(x)^{g(x)}\right]}$

$\ln{\left[h(x)\right]} = g(x)\ln{\left[f(x)\right]}$.

b) $\frac{d}{dx} \left\{\ln{\left[h(x)\right]}\right\}= \frac{d}{dx}\left\{g(x)\ln{\left[f(x)\right]}\right\}$

$= g(x)\,\frac{d}{dx}\left\{\ln{\left[f(x)\right]}\right\} + \ln{\left[f(x)\right]}\,\frac{d}{dx}[g(x)]$

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

c) $y = \ln{\left[h(x)\right]}$.

Let $u = h(x)$ so that $y = \ln{u}$.

$\frac{du}{dx} = h'(x)$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{h(x)}$.

$\frac{dy}{dx} = \frac{h'(x)}{h(x)}$.

Therefore $\frac{d}{dx}\left\{\ln{\left[h(x)\right]}\right\} = \frac{h'(x)}{h(x)}$.

d) Since $\frac{d}{dx}\left\{\ln{\left[h(x)\right]}\right\} = \frac{h'(x)}{h(x)}$ and $\frac{d}{dx} \left\{\ln{\left[h(x)\right]}\right\}= \frac{g(x)f'(x)}{f(x)} + g'(x)\ln{\left[f(x)\right]}$

This means

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

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

And since $h(x) = f(x)^{g(x)}$ this means

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

e) If $h(x) = x^x$

Let $f(x) = x$ so that $f'(x) = 1$.

Let $g(x) = x$ so that $g'(x) = 1$.

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

$= x^x\left[\frac{x\cdot 1}{x} + 1\cdot \ln{x}\right]$

$= x^x\left(1 + \ln{x}\right)$.