Logarithmic differentiation

**Lippi86** · February 22nd 2010, 10:13 PM

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...

**Prove It** · February 22nd 2010, 11:11 PM

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 $g'(x) = 1$ .

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

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

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

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