How do you find the derivative of #ln(x^2+1)#?

1 Answer
Feb 12, 2017

#frac{d}{dx}(y)=frac{2x}{x^2+1}#

Explanation:

Given equation is #y=ln(x^2+1)#

One fundamental equation of derivative calculus is
#frac{d}{dx}(y)=1/x# when #y=lnx#

Another important aspect of derivative calculus is the chain rule, that is if #y=f(t)# and #t=g(x)#, then
#dy/dx=dy/dt*dt/dx#

Taking #t=x^2+1#, from chain rule

#dy/dx=\frac{d}{dt}(lnt)*dt/dx#
#\implies dy/dx=1/t*fracddx(x^2+1)#

Substitute for whatever #t# remains in the equation as #x^2+1# and you'll get the answer.