# Math Help - Equation of the tangent to the curve.

1. ## Equation of the tangent to the curve. [Solved]

Find the Equation of the tangent to the curve of e^2y + xy= 4x + 1,

What I've done so far,

e^2y + xy= 4x + 1

e^2y + xy - 4x - 1 = 0

dy/dx = 2ye^2y + 1 - 4 = 0

dy/dx = 2ye^2y - 3 = 0

I'm not sure if I've done it right because there is no x value, does this mean I sub y=0 into dy/dx instead.

Any help would be appreciated.

2. Originally Posted by driverfan2008
Find the Equation of the tangent to the curve of e^2y + xy= 4x + 1,

What I've done so far,

e^2y + xy= 4x + 1

e^2y + xy - 4x - 1 = 0

dy/dx = 2ye^2y + 1 - 4 = 0

dy/dx = 2ye^2y - 3 = 0

I'm not sure if I've done it right because there is no x value, does this mean I sub y=0 into dy/dx instead.

Any help would be appreciated.
This should be in the calculus section, but never mind...

Since this equation can be solved for x, use the formula $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.

$e^{2y} + xy = 4x + 1$

$xy - 4x = 1 - e^{2y}$

$x(y - 4) = 1 - e^{2y}$

$x = \frac{1 - e^{2y}}{y - 4}$

$\frac{dx}{dy} = \frac{(y - 4)\frac{d}{dy}(1 - e^{2y}) - (1 - e^{2y})\frac{d}{dy}(y - 4)}{(y - 4)^2}$

$\frac{dx}{dy} = \frac{e^{2y}(4 - y) + e^{2y} - 1}{(y - 4)^2}$

$\frac{dy}{dx} = \frac{(y - 4)^2}{e^{2y}(4 - y) + e^{2y} - 1}$.

3. Originally Posted by Prove It
This should be in the calculus section, but never mind...

Since this equation can be solved for x, use the formula $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.

$e^{2y} + xy = 4x + 1$

$xy - 4x = 1 - e^{2y}$

$x(y - 4) = 1 - e^{2y}$

$x = \frac{1 - e^{2y}}{y - 4}$

$\frac{dx}{dy} = \frac{(y - 4)\frac{d}{dy}(1 - e^{2y}) - (1 - e^{2y})\frac{d}{dy}(y - 4)}{(y - 4)^2}$

$\frac{dx}{dy} = \frac{e^{2y}(4 - y) + e^{2y} - 1}{(y - 4)^2}$

$\frac{dy}{dx} = \frac{(y - 4)^2}{e^{2y}(4 - y) + e^{2y} - 1}$.
Oh sorry for posting in the wrong area. Many thanks for your help its very appreciated =)