# How do you use implicit differentiation to find (dy)/(dx) given 2x^3=(3xy+1)^2?

Sep 27, 2016

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

#### Explanation:

$2 {x}^{3} = {\left(3 x y + 1\right)}^{2}$.

Diff.ing both sides w.r.t. x,

$\frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right) = \frac{d}{\mathrm{dx}} {\left(3 x y + 1\right)}^{2}$.

Here, by the Chain Rule,

$\text{The R.H.S.=} \frac{d}{\mathrm{dx}} {\left(3 x y + 1\right)}^{2} = 2 \left(3 x y + 1\right) \frac{d}{\mathrm{dx}} \left(3 x y + 1\right)$

$= 2 \left(3 x y + 1\right) \left\{\frac{d}{\mathrm{dx}} \left(3 x y\right) + \frac{d}{\mathrm{dx}} \left(1\right)\right\}$

$= 2 \left(3 x y + 1\right) \left\{3 \frac{d}{\mathrm{dx}} \left(x y\right) + 0\right\}$

$= 2 \left(3 x y + 1\right) \left[3 \left\{x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left(x\right)\right\}\right] \ldots \ldots . . \text{[Product Rule]}$

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

$= 6 \left(3 x y + 1\right) \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$

$\text{The L.H.S.=} \frac{d}{\mathrm{dx}} \left(2 {x}^{3}\right) = 2 \cdot 3 {x}^{2} = 6 {x}^{2.} \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$

From $\left(1\right) \mathmr{and} \left(2\right)$, we have,

$\cancel{6} {x}^{2} = \cancel{6} \left(3 x y + 1\right) \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) \text{, i.e.,}$

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

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

Enjoy Maths.!