Question

How do you find the equation of tangent line to the curve `(2x+3)^(1/2)` at the point x=3?

Answer 1

`y=1/3x+2`

Explanation:

`"the derivative of the function at x = 3 gives the slope"`
`"of the tangent line"`

`"differentiate using the "color(blue)"chain rule"`

`"Given "y=f(g(x))" then "`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`"here "y=(2x+3)^(1/2)`

`rArrdy/dx=1/2(2x+3)^(-1/2)xxd/dx(2x+3)`

`color(white)(rArrdy/dx)=(2x+3)^(-1/2)=1/(2x+3)^(1/2)`

`"at "x=3tody/dx=1/3" and "y=3`

`"using "m=1/3" and "(x_1,y_1)=(3,3)`

`y-3=1/3(x-3)larrcolor(blue)"point-slope form"`

`y-3=1/3x-1`

`rArry-1/3x+2larrcolor(red)"equation of tangent"`