# Logarithmic Differentiation

• Dec 13th 2009, 01:01 PM
thekrown
Logarithmic Differentiation
4. e) Use Logarithmic differention to find the derivative of the function.

f(x) = (cos x)^x

I can turn it into x ln cos x but I do not know how to finish this. Please help.
• Dec 13th 2009, 01:10 PM
Also sprach Zarathustra
f(x) = (cos x)^x

lnf(x)=ln{(cos x)}^x

lnf(x)=x*ln(cos x)

Now we derivative according to x.

f'(x)/f(x) = ln(cos x) + x*{(-sin x)/(cos x))

f'(x) = f(x)*{ln(cos x) + x*{(-sin x)/(cos x))}

f'(x)= {(cos x)^x}*{ln(cos x) + x*{(-sin x)/(cos x))}
• Dec 13th 2009, 01:25 PM
makaan
$y=(\cos x)^x$

Take the ln of both sides.

$ln(y)=ln(\cos x)^x=x \cdot ln(\cos x)$

Differentiate both sides with respect to x.

$\frac{d}{dx}[ln(y)]=\frac{d}{dx}[x \cdot ln(\cos x)]$

Use the chain rule on the left hand side and the product rule on the RHS.

$\frac{1}{y} \cdot \frac{dy}{dx}= ln(\cos x)+x\frac{d}{dx}[ln(\cos x)]$

Now use the chain rule on the RHS.

$\frac{1}{y} \cdot \frac{dy}{dx}=ln(\cos x)+x (\frac{1}{\cos x}) (-\sin x)=ln(\cos x)-x \tan x$

Finally multiply both sides by 'y'.

$\frac{dy}{dx}=y[ln(\cos x)-x \tan x]=(\cos x)^x[ln(\cos x)-x \tan x]$.