Question

Differentiate the function given below w.r.t. $x:$

$\frac{3x+4}{5x^2-7x+9}$

Answer 1

Let $y=\frac{3x+4}{5x^2-7x+9}$

$\Rightarrow \frac{dy}{dx}=\frac{\left(\begin{array}{c}(5x^2-7x+9)\frac{d(3x+4)}{dx}-\\ (3x+4).\frac{d}{dx}(5x^2-7x+9)\end{array}\right)}{(5x^2-7x+9{)}^2}$

$[\because \frac{d\left(\frac{u}{v}\right)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}]$

$\Rightarrow \frac{dy}{dx}=\frac{\left(\begin{array}{c}(5x^2-7x+9)[\frac{d\left(3x\right)}{dx}+\frac{d\left(4\right)}{dx}]-\\ (3x+4)[\frac{d\left(5x^2\right)}{dx}-\frac{d\left(7x\right)}{dx}+\frac{d\left(9\right)}{dx}]\end{array}\right)/}{(5x^2-7x+9{)}^2}$

$[\because \frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx}]$

$\Rightarrow \frac{dy}{dx}=\frac{\left(\begin{array}{c}(5x^2-7x+9)\times 3-\\ (3x+4)(10x-7)\end{array}\right)}{(5x^2-7x+9{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{\left(\begin{array}{c}15x^2-21x+27\\ -30x^2+21x-40x+28\end{array}\right)}{(5x^2-7x+9{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{55-40x-15x^2}{(5x^2-7x+9{)}^2}$