Question

Pls show solution tysm

Answer 1

Given:

The equation of curve is:

`y=(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=(dy)/(dx)_((a,b))`

We have,

`y=(2e^x)/(1+e^x)`

Differentiate with respect to x.

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

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

Slope of the tangent is

`(dy)/(dx)_((0,1))=((1+e^0)(2e^0)-2e^0(e^0))/((1+e^0)^2)`

`(dy)/(dx)_((0,1))=((1+1)(2(1))-2(1)(1))/((1+1)^2)`

`(dy)/(dx)_((0,1))=((2)(2)-2)/((2)^2)`

`(dy)/(dx)_((0,1))=(4-2)/(4)`

on further simplification, we get

`(dy)/(dx)_((0,1))=(2)/(4)`

`(dy)/(dx)_((0,1))=(1)/(2)`

The slope of the tangent line is
`m=(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=(1)/(2)(x-0)`

`y-1=(1)/(2)x`

`y=(1)/(2)x+1`

Therefore, the equation of the tangent line is
`y=(1)/(2)x+1`.