# How do you use implicit differentiation to find the slope of the curve given x^2y=x+2 at (2,1)?

Oct 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{4}$

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

The $\textcolor{b l u e}{\text{product rule}}$ is required to differentiate ${x}^{2} y$

${x}^{2.} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y = 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y}{x} ^ 2$

Use the point (2 ,1) to evaluate $\frac{\mathrm{dy}}{\mathrm{dx}}$ which gives the slope.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \left(2 \times 2 \times 1\right)}{2} ^ 2 = - \frac{3}{4}$

Oct 21, 2016

Please see the explanation for the procedure. The slope is $- \frac{3}{4}$

#### Explanation:

I will explain each term.

Term 1: $\frac{d \left[{x}^{2} y\right]}{\mathrm{dx}}$ requires the use of the product rule.

$\frac{d \left[g \left(x\right) h \left(y\right)\right]}{\mathrm{dx}} = g ' \left(x\right) h \left(y\right) + g \left(x\right) h ' \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$g \left(x\right) = {x}^{2} , g ' \left(x\right) = 2 x , h \left(y\right) = y , \mathmr{and} h ' \left(y\right) = 1$

$\frac{d \left[{x}^{2} y\right]}{\mathrm{dx}} = 2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

Term 2:

$\frac{d \left[x\right]}{\mathrm{dx}} = 1$

Term 3:

$\frac{d \left[2\right]}{\mathrm{dx}} = 0$

Put these back into their corresponding locations in the equation:

$2 x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + 0$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

${x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x y + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x y}{x} ^ 2$

The slope, m, is the above evaluated at the point $\left(2 , 1\right)$

$m = \frac{1 - 2 \left(2\right) \left(1\right)}{2} ^ 2$

$m = - \frac{3}{4}$