Question

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

Answer 1

`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)`