If G(x)= x/(1+2x), find G'(a). Use it to find an equation of the tangent line to the curve ...?

Question

If G(x)= x/(1+2x), find G'(a). Use it to find an equation of the tangent line to the curve y=x/(1+2x) at the point (-1/4, -1/2)

Answer 1

Use quotient rule:

if G(x) = A(x)/B(x),
then G'(x) = [A'(x)B(x) - A(x)B'(x)] / [B(x)^2]

In this case,
A(x) = x, A'(x) = 1
B(x) = 1 + 2x, B'(x) = 2

so G'(x) = [1*(1 + 2x) - x*2] / [(1+2x)^2]
... = 1/(1 + 2x)^2

and G'(a) = 1/(1 + 2a)^2

In the point (-1/4, -1/2), the derivative is

G'(-1/4) = 1/(1 - 2/4)^2 = 4

so the line has equation

y = 4 x + b

using x = -1/4, y = -1/2, you get b = 1/2:

y = 4 x + 1/2

Answer 2

Taking Calc I are we? Well, find the derivative G'(a). You probably are doing this by the limit definition. Then plug in a=-1/4 to find the slope of the tangent line. Then find the equation of the line with that slope that goes through (-1/4, -1/2).

Answer 3

When you find G'(a), you have the slope of the tangent line.

G'(x) = [(1 + 2x)*1 - x*2] / (1 + 2x)^2
=1 / (4x^2 + 4x + 1)

G'(a) = 1 / (4a^2 + 4a + 1)

To evaluate, plug in the x value given in (-1/4 , -1/2)

G'(-1/4) = 1 / (4(-1/4)^2 + 4(-1/4) + 1)
=1 / [(1/4) - 1 + 1] = 4

m = 4 = slope of the tangent line.

Now, use the slope and the given point in the point-slope equation:

y - (-1/2) = 4(x - (-1/4))
y + (1/2) = 4x + 1
y = 4x - (1/2)

Answer 4

am going to apply the product rule with the help of bringing the backside to the best: =x*(a million+2x)^-a million D1*2+D2*a million D1=a million D2=-a million*(a million+2x)^-2*2 simplified: -2*(a million+2x)^-2 <----- I used the chain rule now plugging the stuff in: [a million/(a million+2x)] +[ -25123cdf9e54ddece6db7cc6bef4d31+2x)^2 ] now basically plug interior the factor and you're carried out

Answer 5

i dont know

Answer 6

=You do your own homework