# Thread: Help with logarithmic differention

1. ## Help with logarithmic differention

Hey I need help with this problem:

Solve using logarithmic differentiation:
y=((x^2+1)/(x^2-1))^(1/4)

Thanks!

2. Hello

$ln y = \frac{1}{4} ln \frac{x^2 +1}{x^2-1}$

$ln y = \frac{1}{4} ln (x^2 +1) - \frac{1}{4} ln (x^2-1)$

differentiating w.r.t.x

$\frac{\frac{dy}{dx}}{y} ln y = \frac{1}{4} ( \frac{2x}{x^2+1} - \frac{2x}{x^2 -1} )$

= $\frac{1}{2} ( \frac{x}{x^2 +1} - \frac{x}{x^2 -1} )$

therefore

$\frac{dy}{dx} = \frac{1}{2} \left( \frac{x^2 +1}{x^2 -1} \right)^{\frac{1}{4}} \left( \frac{x}{x^2 +1} - \frac{x}{x^2 -1} \right)$

3. Originally Posted by luckeyduck
Hey I need help with this problem:

Solve using logarithmic differentiation:
y=((x^2+1)/(x^2-1))^(1/4)

Thanks!
first log both sides:

$\Rightarrow \ln y = \ln \left[ \frac {x^2 + 1}{(x + 1)(x - 1)} \right]^{1/4}$

$\Rightarrow \ln y = \frac 14 \ln (x^2 + 1) - \frac 14 \ln (x + 1) - \frac 14 \ln (x - 1)$

now differentiate implicitly