# How do you find the equation of the tangent line to the curve y = (2e^x)/(1+e^x) at the pt (0,1)?

Feb 18, 2017

graph{(y - (2e^x)/(1+e^x)) (y- (1/2x + 1)) = 0 [-10, 10, -5.21, 5.21]}

The equation of the tangent line at $\left(0 , 1\right)$ is $y = \frac{1}{2} x + 1$

#### Explanation:

Consider $x = 0 , y = 1$

Using Quotient Rule of Differentiation:

Let $u = 2 {e}^{x}$ and $v = 1 + {e}^{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v u ' - u v '}{v} ^ 2$, and given that $d {e}^{x} / \mathrm{dx} = {e}^{x}$

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

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

dy/dx = ((2e^(x) )/ (1+e^x)) - ((2e^x)(e^(x)))/ (1+e^x)^2)

When $x = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[\frac{2 {e}^{0}}{1 + {e}^{0}}\right] - \left\{\frac{\left[\left(2 {e}^{0}\right)\right] \left[{e}^{0}\right]}{{\left(1 + {e}^{0}\right)}^{2}}\right\}$

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2}$

So the slope at $x = 0$ is $\frac{1}{2}$ $\left(f ' \left(x\right) = m = \frac{1}{2}\right)$.

Consider:

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

${y}_{0} = 1 , {x}_{0} = 0$ ( this is at the point $\left(0 , 1\right)$ )

$y - f \left({x}_{0}\right) = f ' \left({x}_{0}\right) \left(x - {x}_{0}\right)$

$y - 1 = \frac{1}{2} \left(x - 0\right)$

$y = \frac{1}{2} x + 1$

So the equation of the tangent line at $\left(0 , 1\right)$ is $y = \frac{1}{2} x + 1$