# What is the slope of the tangent line of  (x-y^2)/(3xe^(y^2)) =C , where C is an arbitrary constant, at (1,5)?

Oct 22, 2016

The slope, $m = \frac{5}{46}$

#### Explanation:

Specifying the point $\left(1 , 5\right)$ makes C become a defined constant:

$C = \frac{1 - {5}^{2}}{3 \left(1\right) \left({e}^{{5}^{2}}\right)} = \frac{- 24}{3 {e}^{25}} = - 8 {e}^{- 25}$

However, this has no bearing on the slope of the tangent line, because it will be 0, when we differentiate:

$\frac{d \left(- 8 {e}^{- 25}\right)}{\mathrm{dx}} = 0$

Use the Quotient Rule to differentiate the left side:

$f \left(x , y\right) = g \frac{x , y}{h \left(x , y\right)}$, then $f ' \left(x , y\right) = \frac{g ' \left(x , y\right) h \left(x , y\right) - g \left(x , y\right) h ' \left(x , y\right)}{h \left(x , y\right)} ^ 2$

Let $g \left(x , y\right) = x - {y}^{2}$, then $g ' \left(x , y\right) = 1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$
Let $h \left(x , y\right) = 3 x {e}^{{y}^{2}}$, then $h ' \left(x , y\right) = 3 {e}^{{y}^{2}} + 6 x y {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}$

This is equal to 0, therefore, we can rid ourselves of the denominator by multiplying both sides by it.

Substitute the above into the numerator and set it equal to 0:

$\left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(3 x {e}^{{y}^{2}}\right) - \left(x - {y}^{2}\right) \left(3 {e}^{{y}^{2}} + 6 x y {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(3 x {e}^{{y}^{2}}\right) + \left({y}^{2} - x\right) \left(3 {e}^{{y}^{2}} + 6 x y {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

3xe^(y^2) - 6xye^(y^2)dy/dx + 3(y^2 - x)e^(y^2) + 6xy(y^2 - x)e^(y^2)dy/dx) = 0

$3 {y}^{2} {e}^{{y}^{2}} - 6 x y {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 x y \left({y}^{2} - x\right) {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$3 {y}^{2} {e}^{{y}^{2}} + 6 x y \left({y}^{2} - x - 1\right) {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$6 x y \left({y}^{2} - x - 1\right) {e}^{{y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {y}^{2} {e}^{{y}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {y}^{2} {e}^{{y}^{2}}}{6 x y \left({y}^{2} - x - 1\right) {e}^{{y}^{2}}}$

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

The slope, m, is the above evaluated at the point $\left(1 , 5\right)$:

$m = \frac{5}{2 \left(1\right) \left({5}^{2} - 1 - 1\right)}$

$m = \frac{5}{2 \left(23\right)}$

$m = \frac{5}{46}$