# Thread: Hello, need help with finding equation for a tangent line with the given function

1. ## Hello, need help with finding equation for a tangent line with the given function

Hello! Thank you for your time and I really need help with finding the equation of the tangent line for 1+ln2xy=e^(2x-y) at the point (1/2,1)

If someone would please help me and if they could add steps to it as well so that I can have a better understanding. Its probably simple and I'm just overthinking but I could really use some help

2. ## Re: Hello, need help with finding equation for a tangent line with the given function

Originally Posted by Sonofrhot6
Hello! Thank you for your time and I really need help with finding the equation of the tangent line for 1+ln2xy=e^(2x-y) at the point (1/2,1)

If someone would please help me and if they could add steps to it as well so that I can have a better understanding. Its probably simple and I'm just overthinking but I could really use some help
In a few words: take the implicit derivative of both sides:
$\displaystyle \frac{1}{2xy} \cdot \left ( 2y + 2x \cdot \frac{dy}{dx} \right ) = \left ( 2 - \frac{dy}{dx} \right ) \cdot e^{2x - y}$

Usually I would recommend solving for dy/dx first, but in this case it may be simpler for you to put in the coordinates of the point, then solve for dy/dx.

-Dan

3. ## Re: Hello, need help with finding equation for a tangent line with the given function

$1+\ln(2xy) = e^{2x-y}$

differentiate by $x$ on each side, $\dfrac{dy}{dx} = y^\prime$

$\dfrac{2y+2xy^\prime}{2xy} = e^{2x-y}(2-y^\prime)$

Solve for $y^\prime$

$y^\prime = -\dfrac{y \left(e^y-2 e^{2 x} x\right)}{x \left(e^{2 x} y+e^y\right)}$

now just plug your point $\left(\dfrac 1 2,~1\right)$ and evaluate

I imagine your prof will want more details. I leave it to you to provide them.