Consider the function y=x/(x+1)

a) Find dy/dx

b) Find the equation of the line tangent to the curve at x=-2

c) Find the equation of the line normal to the curve at x=1

dy/dx = [(x+1)1 -x(1) ] / (x+1)^2

= 1/(x+1)^2

at x = -2
y = -2/-1 = 2 so through point (-2,2)
dy/dx = m = 1/1 = 1
so
y = 1 x + b
2 = -2 + b
b = 4
y = x +
for part c
find a new m at x = 1, y = 1/2
the m we want = -1/m
then repeat method of part b

a) To find the derivative, dy/dx, of the function y = x/(x+1), we can use the quotient rule. The quotient rule states that if you have a function of the form u/v, where u and v are functions of x, the derivative is given by:

dy/dx = (v * du/dx - u * dv/dx) / v^2

In this case, u = x and v = x+1. Let's find the derivatives of u and v:

du/dx = 1 (since the derivative of x with respect to x is 1)

dv/dx = 1 (since the derivative of a constant function, in this case, 1, with respect to x is 0)

Now we can substitute these values into the quotient rule formula:

dy/dx = [(x+1)(1) - x(1)] / (x+1)^2

Simplifying further:

dy/dx = (x+1 - x) / (x+1)^2

dy/dx = 1 / (x+1)^2

So the derivative of y with respect to x, or dy/dx, is 1 / (x+1)^2.

b) To find the equation of the tangent line to the curve at x = -2, we need to find the slope of the tangent line and a point on the line.

First, let's find the slope of the tangent line by substituting x = -2 into the derivative we found above:

slope = dy/dx = 1 / ((-2)+1)^2 = 1 / (-1)^2 = 1

The slope of the tangent line is 1.

Now, let's find a point on the line. We can do this by substituting x = -2 into the original function y = x/(x+1):

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

Therefore, the point on the line is (-2, 2).

Using the slope (1) and the point (-2, 2), we can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is the point:

y - 2 = 1(x - (-2))

y - 2 = x + 2

Collecting terms:

y = x + 4

So, the equation of the line tangent to the curve at x = -2 is y = x + 4.

c) To find the equation of the line normal to the curve at x = 1, we need to find the slope of the normal line and a point on the line.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. In this case, the slope of the tangent line is 1, so the slope of the normal line will be -1.

Now, let's find a point on the line. We can do this by substituting x = 1 into the original function y = x/(x+1):

y = 1/(1+1) = 1/2

Therefore, the point on the line is (1, 1/2).

Using the slope (-1) and the point (1, 1/2), we can use the point-slope form of a line to find the equation of the normal line:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is the point:

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

y - 1/2 = -x + 1

Collecting terms:

y = -x + 3/2

So, the equation of the line normal to the curve at x = 1 is y = -x + 3/2.