Question

Find the derivative of the fuctiong ((1)/((x^2-3)(x^2+3x+5)))

Answer 1

\[1/(x^2-3)(x^2+3x+5)\]

Answer 2

f(x) = 1/[(x^2-3)(x^2+3x+5)] = (x^2-3)^(-1) (x^2+3x+5)^(-1)

f'(x) = -(x^2-3)^-2 * (2x) *(x^2+3x+5)^(-1) - (x^2+3x+5)^(-2) * (2x+3)*(x^2-3)^(-1)

= (x^2-3)^(-2)*(x^2+3x+5)^(-2) * [-2x(x^2+3x+5)-(2x+3)(x^2-3)]

Answer 3

\[=\frac{ -(2x)(x^2+3x+5)-(2x+3)(x^2-3) }{ (x^2-3)^{2}(x^2+3x+5)^{2} }\]

Answer 4

Thank you, I tried a little bit and got the answer I just thought it was harder than it was and didn't want to put in the effort lol.....

Answer 5

Refer to the attachment.