Differentiate using the Quotient Rule –
[tex]\qquad[/tex][tex]\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\[/tex]
According to the given question, we have –
- f(x) = x^3+5x+2
- g(x) = x^2-1
Let's solve it!
[tex]\qquad[/tex][tex]\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\[/tex]
[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\[/tex]
[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\[/tex]
[tex]\qquad[/tex][tex] \pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\[/tex]
[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\[/tex]
[tex]\qquad[/tex][tex]\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\[/tex]
[tex]\qquad[/tex][tex]\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\[/tex]
[tex]\qquad[/tex][tex]\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\[/tex]
