another derivative..

**tbenne3** · February 24th 2010, 11:27 AM

will someone do a step by step so I can try and teach myself this stuff..

Let

**icemanfan** · February 24th 2010, 11:41 AM

$\frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2}$

The chain rule:
$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

Let $f(x) = \tan^{-1} x, g(x) = \cos (7x)$

$\frac{d}{dx} \tan^{-1} (\cos (7x)) =$

$\frac{1}{1 + (\cos (7x))^2} \cdot \frac{d}{dx} \cos (7x) =$

$\frac{-\sin (7x)}{1 + (\cos (7x))^2} \cdot \frac{d}{dx} (7x) =$

$\frac {-7 \sin (7x)}{1 + (\cos (7x))^2}$

**tbenne3** · February 24th 2010, 11:52 AM

thank you very much

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