Question

Find dy/dx for (3x^2-2/2x+3)^5... chain rule is used to find its derivative but I’m having trouble. Need by 3pm, please help!

Answer 1

`f(x)=\left((3x^2-2)/(2x+3)\right)^5\\\\\text{Use:}\\\left((f(x))/(g(x))\right)=(f'(x)g(x)-f(x)g'(x))/([g(x)]^2)\\\\\left([f(x)]^n\right)'=n\left[f(x)\right]^(n-1)\cdot f'(x)`



`f'(x)=5\cdot\left((3x^2-2)/(2x+3)\right)^4\cdot((3x^2-2)'(2x+3)-(3x^2-2)(2x+3)')/((2x+3)^2)\\\\=5\cdot\left((3x^2-2)/(2x+3)\right)^4\cdot(6x(2x+3)-(3x^2-2)\cdot2)/((2x+3)^2)\\\\=5\cdot\left((3x^2-2)/(2x+3)\right)^4\cdot(12x^2+18x-6x^2+4)/((2x+3)^2)\\\\=5\cdot\left((3x^2-2)/(2x+3)\right)^4\cdot(6x^2+18x+4)/((2x+3)^2)`