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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!

Respuesta :

gmany
[tex]f(x)=\left(\dfrac{3x^2-2}{2x+3}\right)^5\\\\\text{Use:}\\\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{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)[/tex]


[tex]f'(x)=5\cdot\left(\dfrac{3x^2-2}{2x+3}\right)^4\cdot\dfrac{(3x^2-2)'(2x+3)-(3x^2-2)(2x+3)'}{(2x+3)^2}\\\\=5\cdot\left(\dfrac{3x^2-2}{2x+3}\right)^4\cdot\dfrac{6x(2x+3)-(3x^2-2)\cdot2}{(2x+3)^2}\\\\=5\cdot\left(\dfrac{3x^2-2}{2x+3}\right)^4\cdot\dfrac{12x^2+18x-6x^2+4}{(2x+3)^2}\\\\=5\cdot\left(\dfrac{3x^2-2}{2x+3}\right)^4\cdot\dfrac{6x^2+18x+4}{(2x+3)^2} [/tex]

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