How do you find the second derivative of $ln(x^2+10)$ ?

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Answer 1 · 2016-11-21T19:06:46

$(d^2y)/(dx^2) = (2(10 - x^2))/(x^2 + 10)^2$

We find the first derivative , and then differentiate again.

$y = ln(x^2 + 10)$

$e^y = x^2 + 10$

$e^y(dy/dx) = 2x$

$dy/dx= ( 2x)/(e^(y)$

$dy/dx= (2x)/(e^ln(x^2 + 10)$

$dy/dx = (2x)/(x^2 + 10)$

$(d^2y)/(dx^2) = (2(x^2 + 10) - 2x(2x))/(x^2 + 10)^2$

$(d^2y)/(dx^2) = (2x^2 + 20 - 4x^2)/(x^2 + 10)^2$

$(d^2y)/(dx^2) = (2(10 - x^2))/(x^2 + 10)^2$

Hopefully this helps!

How do you find the second derivative of ln(x^2+10) ln(x^2+10) ?