If g(x)=x/1+2x, how do you find g'(a) and use it to find an equation of the tangent line to the curve at the point (1, 1/3).

equation of tangent at (1,1/3) is:

Step-by-step explanation:

this differentiation can be easily done using the quotient rule:

here,

u = x, and v = 1+2x.

u' = 1, and v' = 2

The derivative of u and v are written as u' and v'.

this is the derivative of g(x) and it is denoted by g'(x)

g'(a) means to simply replace all the x's with a in the above equation

to find the equation of the tangent to this curve at (1,1/3)

we need to know the slope of the tangent at that very point. This can be found using g'(x). since g'(x) only takes x as the input. we'll only use the x-coordinate of the point (1,1/3)

the slope of the tangent to the curve at the point (1,1/3) is 1/9

to find the equation of the line we'll use:

here m is the slope, and (x1,y1) = (1,1/3)

simplify:

this is the equation of the tangent to the curve g(x) at the point (1,1/3)

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363