Question

Y equals the quotient of the quantity x squared plus 4 times x and the quantity x cubed minus 5.

Answer 1

Given:

Consider the completer question is "Find the derivative $\dfrac{dy}{dx}$ for $y=\dfrac{x^2-4x}{x^3-5}$."

To find:

The derivative $\dfrac{dy}{dx}$.

Solution:

Chain rule: $\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Quotient rule: $\dfrac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$

We have,

$y=\dfrac{x^2-4x}{x^3-5}$

Differentiate with respect to x.

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\dfrac{x^2-4x}{x^3-5}\right)$

Using chain rule and quotient rule, we get

$\dfrac{dy}{dx}=\dfrac{(x^3-5)\dfrac{d}{dx}(x^2-4x)-(x^2-4x)\dfrac{d}{dx}(x^3-5)}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{(x^3-5)(2x-4)-(x^2-4x)(3x^2)}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{2x^4-4x^3-10x+20-3x^4+12x^3}{(x^3-5)^2}$

$\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}$

Therefore, the required answer is $\dfrac{dy}{dx}=\dfrac{-x^4+8x^3-10x+20}{(x^3-5)^2}$.