What is the derivative of #x tan^-1 - ln sqrt(1+x^2)#?

Question

Cameron Alexander · Accepted Answer

#d/dx(xtan^-1(x)-lnsqrt(1+x^2))=tan^-1(x)# #y=xtan^-1(x)-lnsqrt(1+x^2)# First rewrite the logarithm using #ln(a^b)=bln(a)#: #y=xtan^-1(x)-1/2ln(1+x^2)# Now when we differentiate, we will use the product rule for #xtan^-1(x)# and the chain rule for #ln(1+x^2)#. #dy/dx=(d/dxx)tan^-1(x)+x(d/dxtan^-1(x))-1/2(1/(1+x^2))(d/dx(1+x^2))# #dy/dx=tan^-1(x)+x(1/(1+x^2))-1/2(1/(1+x^2))(2x)# #dy/dx=tan^-1(x)+x/(1+x^2)-x/(1+x^2)# #dy/dx=tan^-1(x)#

Luke Alvarez · Answer

undefined To find the derivative of the given function, $f(x) = x\tan^{-1}(x) - \ln(\sqrt{1+x^2})$, we'll differentiate each term separately and then apply the sum rule for derivatives.

1. $x\tan^{-1}(x)$:
   $$
   \frac{d}{dx}(x\tan^{-1}(x)) = \tan^{-1}(x) + x\left(\frac{1}{1+x^2}\right)
   $$

2. $\ln(\sqrt{1+x^2})$:
   $$
   \frac{d}{dx}\left(\ln(\sqrt{1+x^2})\right) = \frac{1}{\sqrt{1+x^2}} \cdot \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{1+x^2\sqrt{1+x^2}}
   $$

Now, applying the sum rule, we get:
$$
\frac{d}{dx}(f(x)) = \tan^{-1}(x) + x\left(\frac{1}{1+x^2}\right) - \frac{x}{1+x^2\sqrt{1+x^2}}
$$