Derivative of Logarithmic Function

**joatmon** · April 10th 2011, 11:42 AM

Differentiate $h(x) = \ln{(x + \sqrt{x^2-6})}$

My work:

let $u = x + \sqrt{x^2-6}$ then $y = \ln{u}$

$\displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx } = \frac{1}{x + \sqrt{x^2-6}}(1+\frac{1}{2}(x^2-6)^{-1/2}(2x))$

$\displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}$

$\displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}$

$\displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}$

Where did I go wrong??? Thanks.

**Chris L T521** · April 10th 2011, 11:52 AM

Originally Posted by joatmon

Differentiate $h(x) = \ln{(x + \sqrt{x^2-6})}$

My work:

let $u = x + \sqrt{x^2-6}$ then $y = \ln{u}$

$\displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx } = \frac{1}{x + \sqrt{x^2-6}}(\boxed{1}+\frac{1}{2}(x^2-6)^{-1/2}(2x))$

$\displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}$

$\displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}$

$\displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}$

Where did I go wrong??? Thanks.

You forgot about the term I boxed in your work.

The final answer should be $\dfrac{1}{x + \sqrt{x^2-6}}+\dfrac{x}{x\sqrt{x^2-6} + x^2-6}=\dfrac{x+\sqrt{x^2-6}}{x\sqrt{x^2-6}+x^2-6}=\dfrac{1}{\sqrt{x^2-6}}$

I hope this makes sense.

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