How do I find the derivative of #y = log (x^2 + 1)#?

Question

10902 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-08T13:22:07

#dy/dx=(2xlog e)/(x^2+1)#

The derivative of #y=log(u)# is #dy/dx=log e cdot 1/u xx d/dx(u)#, since #lnx = 1/(log e) log x#, and #(d(lnx))/(dx) = 1/x#.

Following these exact steps we get:

#dy/dx=log e cdot (1)/(x^2+1) xx 2x#

The final answer is:

#dy/dx=(2xlog e)/(x^2+1)#

Answer 2 · 2017-06-09T16:58:55

#dy/dx=(2x)/((x^2+1)ln10)#

#y=log(x^2+1)#

We can't differentiate #log(x^2+1)#, so we must rewrite in such a way that we can. As such, the change of base rule will be required:

#log_a b-=log_c b/log_c a#

#log(x^2+1)-=ln(x^2+1)/ln10#

#y=ln(x^2+1)/ln10#

#dy/dx=1/ln10# #[d/dx(ln(x^2+1))]#

#d/dx(ln(x^2+1))=(d/dx(x^2+1))/(x^2+1)=(2x)/(x^2+1)#

#dy/dx=1/ln10xx(2x)/(x^2+1)=(2x)/((x^2+1)ln10)#