Respuesta :

Answer:

[tex]g'(a)=\dfrac{1}{\left(2 a + 1\right)^{2}}[/tex]

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

[tex]y = \dfrac{1}{9}x+\dfrac{2}{9}\\[/tex]

Step-by-step explanation:

[tex]g(x) = \dfrac{x}{1+2x}[/tex]

[tex]g'(x) = \dfrac{d}{dx}\left(\dfrac{x}{1+2x}\right)[/tex]

this differentiation can be easily done using the quotient rule:

[tex]d\left(\dfrac{u}{v}\right) = \dfrac{vu'-uv'}{v^2}[/tex]

here,

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

u' = 1, and v' = 2

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

[tex]g'(x)=\dfrac{(1+2x)(1)-(x)(2)}{(1+2x)^2}\\g'(x)=\dfrac{1}{\left(2 x + 1\right)^{2}}[/tex]

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

[tex]g'(a)=\dfrac{1}{\left(2 a + 1\right)^{2}}[/tex]

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)

[tex]g'(1)=\dfrac{1}{(2(1) + 1)^{2}}[/tex]

[tex]g'(1)=\dfrac{1}{9}[/tex]

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:

[tex](y-y_1) = m(x-x_1)\\[/tex]

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

[tex](y-\frac{1}{3}) = \frac{1}{9}(x-1)\\[/tex]

simplify:

[tex]y = \dfrac{1}{9}x-\dfra{1}{9}+\dfrac{1}{3}\\[/tex]

[tex]y = \dfrac{1}{9}x+\dfrac{2}{9}\\[/tex]

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