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

1 Answer
mason m
Feb 5, 2017

#d/dx(xtan^-1(x)-lnsqrt(1+x^2))=tan^-1(x)#

Explanation:

#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)#

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