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

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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:

$log_a b-=log_c b/log_c a$

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

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

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

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

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

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