# y=x/(2x-3) then find (d^2y)/(dx^2)?

May 3, 2018

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{12}{2 x - 3} ^ 3$

#### Explanation:

.

$y = \frac{x}{2 x - 3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 x - 3\right) \left(1\right) - x \left(2\right)}{2 x - 3} ^ 2 = \frac{2 x - 3 - 2 x}{2 x - 3} ^ 2 = \frac{- 3}{2 x - 3} ^ 2$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{\left(2 x - 3\right)}^{2} \left(0\right) - \left(- 3\right) \left(2\right) \left(2 x - 3\right) \left(2\right)}{2 x - 3} ^ 4 = \frac{12 \left(2 x - 3\right)}{2 x - 3} ^ 4 = \frac{12}{2 x - 3} ^ 3$

May 3, 2018

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{12}{{\left(2 x - 3\right)}^{3}}$

#### Explanation:

Here,

$y = \frac{x}{2 x - 3}$

$\text{Using "color(blue) "Quotient Rule:}$ ( Diff.w.r.t. $x$.)

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 x - 3\right) \frac{d}{\mathrm{dx}} \left(x\right) - x \frac{d}{\mathrm{dx}} \left(2 x - 3\right)}{{\left(2 x - 3\right)}^{2}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 x - 3\right) \left(1\right) - \left(x\right) \left(2\right)}{{\left(2 x - 3\right)}^{2}}$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x - 3 - 2 x}{{\left(2 x - 3\right)}^{2}}$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3}{{\left(2 x - 3\right)}^{2}}$

Again $\text{Using "color(blue) "Quotient Rule:}$ ( Diff.w.r.t. $x$.)

(d^2y)/(dx^2)=((2x-3)^2(0)-(-3)(2(2x-3)2))/((2x- 3)^4)

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{3 \times 2 \times 2 \left(2 x - 3\right)}{{\left(2 x - 3\right)}^{4}}$

$\implies \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{12}{{\left(2 x - 3\right)}^{3}}$