the derivative of arctan

• Feb 27th 2010, 01:11 PM
Jones
the derivative of arctan
Hi,

Hm, trying to get my head around this

the derivative of arctan is $\frac{1}{1+x^2}$

So the derivative of $arctan(0.5sinx)$ should be:

$\frac{cosx}{1+\frac{sin^2x}{4}}$ But that's apparently wrong.

Why?
• Feb 27th 2010, 01:45 PM
Gusbob
Quote:

Originally Posted by Jones
Hi,

Hm, trying to get my head around this

the derivative of arctan is $\frac{1}{1+x^2}$

So the derivative of $arctan(0.5sinx)$ should be:

$\frac{cosx}{1+\frac{sin^2x}{4}}$ But that's apparently wrong.

Why?

Hi, be careful when you're using the chain rule.

$\frac{d}{dx} \arctan \left(\frac{1}{2} \sin (x) \right) = \frac{d \arctan (u)}{du} \times \frac{du}{dx}$ where $u = 0.5 \sin (x)$ and $\frac{d \arctan (u)}{du} = \frac{1}{1+u^2}$

The denominator of your answer is correct. But look at what you did to the numerator. The numerator is $\frac{du}{dx} = \frac{d}{dx} 0.5 \sin (x) = 0.5 \cos (x)$ The 0.5 does not dissapear. That is your only mistake.
• Feb 27th 2010, 02:03 PM
Jones
Thanks(Nod)