# Differentiate

Question:

Differentiate

$\left(\frac{a x^{2}+b x+c}{p x^{2}+q x+r}\right)$

Solution:

To find: Differentiation of $\frac{a x^{2}+b x+c}{p x^{2}+q x+r}$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d x^{n}}{d x}=n x^{n-1}$

Let us take $u=\left(a x^{2}+b x+c\right)$ and $v=\left(p x^{2}+q x+r\right)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left[a x^{2}+b x+c\right]}{d x}=2 a x+b$

$v^{\prime}=\frac{d v}{d x}=\frac{d\left(p x^{2}+q x+r\right)}{d x}=2 p x+q$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{a x^{2}+b x+c}{p x^{2}+q x+r}\right]^{\prime}=\frac{(2 a x+b)\left(p x^{2}+q x+r\right)-\left[a x^{2}+b x+c\right][2 p x+q]}{\left(p x^{2}+q x+r\right)^{2}}$

$=$

$\frac{2 a p x^{3}+2 a q x^{2}+2 a x r+b p x^{2}+b q x+b r-\left[2 a p x^{3}+q a x^{2}+2 b p x^{2}+b q x+2 p c x+c q\right]}{\left(p x^{2}+q x+r\right)^{2}}$

$=\frac{(a q-b p) x^{2}+2(r a-p c) x+b r-c p}{\left(p x^{2}+q x+r\right)^{2}}$

Ans $)=\frac{(a q-b p) x^{2}+2(r a-p c) x+b r-c p}{\left(p x^{2}+q x+r\right)^{2}}$