# Let fx)= (4sinx)/(4sinx+6cosx) The equation of the tangent line to y=f(x) at a=pi/2 can be written in the form y=mx+b. Find m and b

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### 1 Answer

Given `f(x)= (4sinx)/(4sinx+6cosx) `

We need to find the equation of the tangent line at the point x= `pi/2`

First we will find the y-coordinate of the point.

`==gt f(pi/2) = (4sin(pi/2))/(4sin(pi/2)+6cos(pi/2))`

==> f(pi/2)= 4/(4+0)= 4/4 = 1

Then, the tangent line is at the point (`pi/2, 1` )

Now we will find the slope.

The slope (m) is the derivative at x = `pi/2`

`==gt f'(x)= ((4sinx)'(4sinx+6cosx) - (4sinx)(4sinx+6cosx)')/(4sinx+6cosx)^2 `

`==gt f'(x)= (4cosx (4sinx+6cosx)- 4sinx(4cosx-6sinx))/(4sinx+6cosx)^2`

`` `==gt f'(x)= (16sinxcosx + 24cos^2 x - 16sinxcosx + 24sin^2 x)/(4sinx+6cosx)^2 `

`==gt f'(x)= 24/(4sinx+6cosx)^2 `

`==gt f'(pi/2) = 24/(4sin(pi/2) + 6cos(pi/2))^2 = 24/(4^2)= 24/16 `

`==gt f'(pi/2) = 3/2`

Then, the slope m= 3/2

==> Now we will find the equation.

`==gt y-y1= m (x-x1) `

`==gt y-1 = 3/2 (x-pi/2) `

`==gt y= 3/2 x - (3pi/4) + 1 `

`==gt y= 3/2 x + (4-3pi)/4`

**==> m= 3/2 and b= (4-3pi)/4**