Question

Pls show solution tysm

Answer 1

Given:

The equation of curve is:

$y=\dfrac{2e^x}{1+e^x}$

To find:

The equation of the tangent line to the given curve at (0,1).

Solution:

We know that the slope of the tangent line at (a,b) is

$m=\dfrac{dy}{dx}_{(a,b)}$

We have,

$y=\dfrac{2e^x}{1+e^x}$

Differentiate with respect to x.

$\dfrac{dy}{dx}=\dfrac{(1+e^x)(2e^x)'-2e^x(1+e^x)'}{(1+e^x)^2}$

$\dfrac{dy}{dx}=\dfrac{(1+e^x)(2e^x)-2e^x(e^x)}{(1+e^x)^2}$

Slope of the tangent is

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{(1+e^0)(2e^0)-2e^0(e^0)}{(1+e^0)^2}$

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{(1+1)(2(1))-2(1)(1)}{(1+1)^2}$

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{(2)(2)-2}{(2)^2}$

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{4-2}{4}$

on further simplification, we get

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{2}{4}$

$\dfrac{dy}{dx}_{(0,1)}=\dfrac{1}{2}$

The slope of the tangent line is $m=\dfrac{1}{2}$ and it passes through the point (0,1). So, the equation of the tangent line is

$y-y_1=m(x-x_1)$

$y-1=\dfrac{1}{2}(x-0)$

$y-1=\dfrac{1}{2}x$

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

Therefore, the equation of the tangent line is $y=\dfrac{1}{2}x+1$.