Given that #p(x)=(3/((2x^2-5)))#, find #p''(x)#?

Question

Use differentiation please....

1158 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T04:49:36

#p''(x) =(72x^2+60)/(2x^2-5)^3#

Hi Loweena.

Here's one way to do this:
I would start by rewriting the equation without the division line:
#p(x)=3*(2x^2-5)^-1#

From that equation, and applying the chain rule, we can find the first derivative:
#p'(x)=-3*(2x^2-5)^-2*4x#
#p'(x)=-12x*(2x^2-5)^-2#

Now we repeat the process, applying the product rule and the chain rule, to get the second derivative:

#p''(x) = -12*(2x^2-5)^-2 + (-12x)(-2)(2x^2-5)^-3*4x#

#p''(x) = -12*(2x^2-5)^-2 + 96x^2*(2x^2-5)^-3#

If you wish, you can simplify this a little bit.
I will factor out #(2x^2-5)^-3#, and simplify what remains:

#p''(x) = (2x^2-5)^-3*[-12*(2x^2-5)^1+96x^2]#

#p''(x) = (2x^2-5)^-3*[-24x^2+60+96x^2]#

#p''(x) =(2x^2-5)^-3*(72x^2+60)#

If you wish, you can also return to the original format:

#p''(x) =(72x^2+60)/(2x^2-5)^3#