How do you find the derivative #f(x)=arctan(x/alpha)#?

Question

Aurora Alvarado · Accepted Answer

# f'(x)=alpha/(x^2+alpha^2)#

When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule.

#y=arctan(x/alpha) <=> tany=x/alpha #

Differentiate Implicitly:

# sec^2ydy/dx = 1/alpha # ..... [1]

Using the #tan"/"sec# identity;

# tan^2y+1 = sec^2y # # :. (x/alpha)^2+1=sec^2y #

Substituting into [1]

# ((x/alpha)^2+1)dy/dx=1/alpha # # :. ((x^2+alpha^2)/alpha^2)dy/dx=1/alpha # # :. dy/dx=1/alpha * alpha^2/(x^2+alpha^2)# # :. dy/dx=alpha/(x^2+alpha^2)#

Axel Alvarez · Answer

undefined To find the derivative of $ f(x) = \arctan\left(\frac{x}{\alpha}\right) $, where $ \alpha $ is a constant, use the chain rule:

$$ f'(x) = \frac{1}{1 + \left(\frac{x}{\alpha}\right)^2} \cdot \frac{1}{\alpha} $$