Question

i will KISS U ON THE MOUTH IF U HURRY PLS HELP find the derivative dy/dx for y=(x^2+4x) / (x^3-5)

Answer 1

Use the quotient rule:

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

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

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{x^4 - 16 x^3 + 10 x + 20}{(x^3-5)^2}$