# Thread: the derivative of arctan

1. ## the derivative of arctan

Hi,

Hm, trying to get my head around this

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

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

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

Why?

2. Originally Posted by Jones
Hi,

Hm, trying to get my head around this

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

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

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

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

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

The denominator of your answer is correct. But look at what you did to the numerator. The numerator is $\displaystyle \frac{du}{dx} = \frac{d}{dx} 0.5 \sin (x) = 0.5 \cos (x)$ The 0.5 does not dissapear. That is your only mistake.

3. Thanks