Question

Find the derivative dy / dx for y = x^3 - 5x / x^2 - 1

Answer 1

$Use\\\\\left(\dfrac{h(x)}{g(x)}\right)'=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}\\\\\text{We have}\\\\h(x)=x^3-5x\to h'(x)=3x^2-5\\\\g(x)=x^2-1\to g'(x)=2x\\\\f(x)=\dfrac{x^3-5x}{x^2-1}\to f'(x)=\left(\dfrac{x^3-5x}{x^2-1}\right)'\\\\=\dfrac{(x^3-5x)'(x^2-1)-(x^3-5x)(x^2-1)'}{(x^2-1)^2}\\\\=\dfrac{(3x^2-5)(x^2-1)-(x^3-5x)(2x)}{(3x^2-5)^2}\\\\=\dfrac{3x^4-3x^2-5x^2+5-2x^4+10x^2}{(x^2-1)^2}\\\\=\dfrac{x^4+2x^2+5}{(x^2-1)^2}\\\\Answer:\ \left(\dfrac{x^3-5x}{x^2-1}\right)'=\dfrac{x^4+2x^2+5}{(x^2-1)^2}$