Given:
Consider the completer question is "Find the derivative [tex]\dfrac{dy}{dx}[/tex] for [tex]y=\dfrac{x^2-4x}{x^3-5}[/tex]."
To find:
The derivative [tex]\dfrac{dy}{dx}[/tex].
Solution:
Chain rule: [tex]\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)[/tex]
Quotient rule: [tex]\dfrac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}[/tex]
We have,
[tex]y=\dfrac{x^2-4x}{x^3-5}[/tex]
Differentiate with respect to x.
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x^2-4x}{x^3-5}\right)[/tex]
Using chain rule and quotient rule, we get
[tex]\dfrac{dy}{dx}=\dfrac{(x^3-5)\dfrac{d}{dx}(x^2-4x)-(x^2-4x)\dfrac{d}{dx}(x^3-5)}{(x^3-5)^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{(x^3-5)(2x-4)-(x^2-4x)(3x^2)}{(x^3-5)^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{2x^4-4x^3-10x+20-3x^4+12x^3}{(x^3-5)^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}[/tex]
Therefore, the required answer is [tex]\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}[/tex].
