OK, I've worked this problem and it's not coming out right. I think it's because I really don't understand where to plug all of the information into the formula. If you could please explain how to work this example problem, it would be appreciated.
Example:
The point P(1, .5) lies on the curve y=x/(1+x). Use the first derivative to find the slope of the line tangent to the curve at P and then write the equation of the line.
Example:
The point P(1, .5) lies on the curve y=x/(1+x). Use the first derivative to find the slope of the line tangent to the curve at P and then write the equation of the line.
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ok take the first derivative of y using the quotient rule, which is: y=u/v then y' = (u'v - uv')/v^2
(1+x - x)/(1+x)^2 = 1/(1+x)^2
ok plug in 1 to find slope on the tangent line (y') at x=1
y' = 1/(1+1)^2 = 1/4
ok now use point slope to write the equation of the line:
y-.5 = (1/4)(x-1)
if you want it in y=mx+b form you get:
y-.5 = 1/4x - 1/4
y = (1/4)x+ 1/4
EDIT:
ok with the formula lim h--> 0 [f(x+h) - f(x)]/h we get:
lim h-->0 [(x+h)/(1+x+h) - x/(1+x)]/h
simplify to get:
lim h-->0 [(x+x^2+hx+h - x -x^2 -hx)/(1+x)(1+x+h)]/h
and..
lim h-->0 [h/(1+x)(1+x+h)]/h
lim h-->0 h/[(h)(1+x)(1+x+h)]
h's cancel and we get:
lim h-->0 1/(1+x)(1+x+h)
plug in x=1 and h=0
=1/(1+1)(1+1+0)
=1/(2)(2)
= 1/4
that is the slope of the lin at x=1
therefore using point slope we get:
y- 1/2 = (1/4)(x-1)
and in y=mx+b form
y= (1/4)x +1/4
(1+x - x)/(1+x)^2 = 1/(1+x)^2
ok plug in 1 to find slope on the tangent line (y') at x=1
y' = 1/(1+1)^2 = 1/4
ok now use point slope to write the equation of the line:
y-.5 = (1/4)(x-1)
if you want it in y=mx+b form you get:
y-.5 = 1/4x - 1/4
y = (1/4)x+ 1/4
EDIT:
ok with the formula lim h--> 0 [f(x+h) - f(x)]/h we get:
lim h-->0 [(x+h)/(1+x+h) - x/(1+x)]/h
simplify to get:
lim h-->0 [(x+x^2+hx+h - x -x^2 -hx)/(1+x)(1+x+h)]/h
and..
lim h-->0 [h/(1+x)(1+x+h)]/h
lim h-->0 h/[(h)(1+x)(1+x+h)]
h's cancel and we get:
lim h-->0 1/(1+x)(1+x+h)
plug in x=1 and h=0
=1/(1+1)(1+1+0)
=1/(2)(2)
= 1/4
that is the slope of the lin at x=1
therefore using point slope we get:
y- 1/2 = (1/4)(x-1)
and in y=mx+b form
y= (1/4)x +1/4
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The slope of a line is found with the FUNDAMENTAL THEOREM OF CALCULUS EQUATION
(f(x + h) - f(x)) / ((x + h) - x), with the limit as h goes to 0
This is just a variation on the slope formula:
(f(x[2]) - f(x[1])) / (x[2] - x[1])
Where x[2] = x[1] + h
or in a more familiar look:
m = (y[2] - y[1]) / (x[2] - x[1])
y = x / (1 + x)
f(x) = x / (1 + x)
f(x + h) = (x + h) / (1 + x + h)
f(x + h) - f(x) =>
(x + h) / (1 + x + h) - x / (1 + x) =>
((1 + x) * (h + x) - x * (1 + x + h)) / ((1 + x + h) * (1 + x)) =>
(h + xh + x + x^2 - x - x^2 - xh) / ((1 + x + h) * (1 + x)) =>
(h) / ((1 + x + h) * (1 + x))
Divide that by (x + h - x) which simplifies to h
(h / ((1 + x + h) * (1 + x))) / h =>
h / (h * (1 + x + h) * (1 + x)) =>
1 / ((1 + x + h) * (1 + x))
h goes to 0
1 / ((1 + x + 0) * (1 + x)) =>
1 / ((1 + x) * (1 + x)) =>
1 / (1 + x)^2
Now, we just plug in our value for x and that will be the slope of the curve at Point P
x = 1
1 / (1 + 1)^2 =>
1 / (2)^2 =>
1 / 4
The slope at point P is (1/4)
(f(x + h) - f(x)) / ((x + h) - x), with the limit as h goes to 0
This is just a variation on the slope formula:
(f(x[2]) - f(x[1])) / (x[2] - x[1])
Where x[2] = x[1] + h
or in a more familiar look:
m = (y[2] - y[1]) / (x[2] - x[1])
y = x / (1 + x)
f(x) = x / (1 + x)
f(x + h) = (x + h) / (1 + x + h)
f(x + h) - f(x) =>
(x + h) / (1 + x + h) - x / (1 + x) =>
((1 + x) * (h + x) - x * (1 + x + h)) / ((1 + x + h) * (1 + x)) =>
(h + xh + x + x^2 - x - x^2 - xh) / ((1 + x + h) * (1 + x)) =>
(h) / ((1 + x + h) * (1 + x))
Divide that by (x + h - x) which simplifies to h
(h / ((1 + x + h) * (1 + x))) / h =>
h / (h * (1 + x + h) * (1 + x)) =>
1 / ((1 + x + h) * (1 + x))
h goes to 0
1 / ((1 + x + 0) * (1 + x)) =>
1 / ((1 + x) * (1 + x)) =>
1 / (1 + x)^2
Now, we just plug in our value for x and that will be the slope of the curve at Point P
x = 1
1 / (1 + 1)^2 =>
1 / (2)^2 =>
1 / 4
The slope at point P is (1/4)