How do I find the derivative of # f(x) = x tan^-1 - ln sqrt(1+x^2)#?

Question

Aurora Alvarado · Accepted Answer

# tan^-1x#.

#f(x)=xtan^-1x-lnsqrt(1+x^2)=xtan^-1x-ln(1+x^2)^(1/2)#. #:. f(x)=xtan^-1x-1/2ln(1+x^2)#. Using the usual rules of diffn., we get, #f'(x)=x*d/dx{tan^-1x}+tan^-1x*d/dx{x}# #-1/2*d/dx{ln(1+x^2)}#, #=x*1/(1+x^2)+tan^-1x-1/2*1/(1+x^2)*d/dx{1+x^2}#, #=x/(1+x^2)+tan^-1x-1/cancel2*1/(1+x^2)*cancel(2)x#, #=cancel(x/(1+x^2))+tan^-1x-cancel(x/(1+x^2))#. # rArr f'(x)=tan^-1x#. Spread the Joy of Maths!

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = x \tan^{-1}(x) - \ln(\sqrt{1+x^2}) $, you would apply the chain rule and the derivative rules for trigonometric and logarithmic functions:

$$ f'(x) = \frac{d}{dx}[x \tan^{-1}(x)] - \frac{d}{dx}[\ln(\sqrt{1+x^2})] $$

Using the chain rule and the derivative of the natural logarithm function, we get:

$$ f'(x) = \left(1 \cdot \tan^{-1}(x) + x \cdot \frac{1}{1+x^2}\right) - \frac{1}{\sqrt{1+x^2}} \cdot \frac{d}{dx}(1+x^2)^{1/2} $$

Simplify and apply the derivative of $ \tan^{-1}(x) $ and $ (1+x^2)^{1/2} $:

$$ f'(x) = \tan^{-1}(x) + \frac{x}{1+x^2} - \frac{x}{\sqrt{1+x^2} \cdot \sqrt{1+x^2}} $$

$$ f'(x) = \tan^{-1}(x) + \frac{x}{1+x^2} - \frac{x}{1+x^2} $$

$$ f'(x) = \tan^{-1}(x) $$

So, the derivative of $ f(x) = x \tan^{-1}(x) - \ln(\sqrt{1+x^2}) $ is $ f'(x) = \tan^{-1}(x) $.