What are the first and second derivatives of #f(x)=1/(x^2+6)#?

Question

928 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-10T23:13:28

#f(x)=1/(x^2+6) => " " f''(x)=(6(x^2-2))/((x^2+6)^3#

Since #1/x=x^(-1)#, we can say

#f(x)=1/(x^2+6)=(x^2+6)^(-1)#

Then, by the chain rule #(f(g(x))'=f'(g(x))g'(x)#

#=> f'(x)=-2x(x^2+6)^(-2)=(-2x)/((x^2+6)^2)#

To get the 2nd derivative, we simply take the derivative of the derivative

Another way to write #f'(x)#

#f'(x)=-2x(x^2+6)^-2#

Then, by the Product Rule #(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)#

#=> f''(x)=-2(x^2+6)^-2+8x^2(x^2+6)^(-3)#

#=(-2(x^2+6))/((x^2+6)^3)+(8x^2)/(x^2+6)^3#

#=(8x^2-2x^2-12)/((x^2+6)^3)=(6x^2-12)/((x^2+6)^3)=(6(x^2-2))/((x^2+6)^3)#