Question

Solve : $\frac{d}{dx}\left(\frac{3x+9}{2-x}\right)$

Answer 1

Answer:

12x/2x

= 6x Is The answer of This Question

Answer 2

$\bf{\dfrac{d}{dx}\left(\dfrac{3x+9}{2-x}\right)=\dfrac{15}{\left(2-x\right)^2}}$

SOLUTION :

$\mathrm{Apply\:the\:Quotient\:Rule}:\quad \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2}$

$=\frac{\frac{d}{dx}\left(3x+9\right)\left(2-x\right)-\frac{d}{dx}\left(2-x\right)\left(3x+9\right)}{\left(2-x\right)^2}$

$\frac{d}{dx}\left(3x+9\right)=3$

$\frac{d}{dx}\left(2-x\right)=-1$

$=\frac{3\left(2-x\right)-\left(-1\right)\left(3x+9\right)}{\left(2-x\right)^2}$

$\mathrm{Simplify\:}\frac{3\left(2-x\right)-\left(-1\right)\left(3x+9\right)}{\left(2-x\right)^2}$

$=\dfrac{15}{\left(2-x\right)^2}$