If F(x) = 5x/(1 + x^2), find F'(2) and use it to find an equation of the tangent line to the curve y = 5x/(1 + x^2) at the point (2, 2).
-
so F'(2) will be your slope, m, in y=mx+b
F'(x)= [5(1+x^2) - 5x(2x)]/[1+x^2]^2 (quotient rule)
so F'(2)= [5(5)-10(4)]/(25) = -15/25 = -3/5
solve for y-int, given (x,y) --> (2,2)
2 = (-3/5)(2) +b
b = 2 +6/5 = 16/5
tangent line = (-3/5)x + 16/5
F'(x)= [5(1+x^2) - 5x(2x)]/[1+x^2]^2 (quotient rule)
so F'(2)= [5(5)-10(4)]/(25) = -15/25 = -3/5
solve for y-int, given (x,y) --> (2,2)
2 = (-3/5)(2) +b
b = 2 +6/5 = 16/5
tangent line = (-3/5)x + 16/5
-
1. Find F'(x).
Think of F(x) in terms of F(x) = 5x * (1 + x^2)^(-1) and you can use the product rule.
Product rule: If f(x) = a(x) * b(x), then f'(x) = a(x)*b'(x) + a'(x)*b(x)
In this case, a(x) = 5x and b(x) = (1 + x^2)^(-1)
a'(x) = 5
b'(x) = -1(1 + x^2)^(-2) * 2x by using the power rule). Simplifying, b'(x) = -2x/(1 + x^2)^2
So F'(x) = 5x * -2x/(1 + x^2)^2 + 5 * 1/(1 + x^2) = -10x^2 / (1 + x^2)^2 + 5 / (1 + x^2)
This could be simplified further, but that is not necessary since you just need to find F'(x) for a specific x = a. (In this case, a = 2.)
2. Evaluate F'(x) at x = 2.
F'(2) = -10 * 2^2 / (1 + 2^2)^2 + 5 / (1 + 2^2)
= -40 / 5^2 + 5 / 5
= -40/25 + 1
= -15/25
= -3/5
This gives you the slope of the line that is tangent to the curve F(x) at the point (2, 2).
3. Now use the slope of the tangent line m = -3/5 at the point (2,2) to find the equation of the tangent line.
y - y1 = m ( x - x1) where your slope is m and your point is (x1, y1)
If you are taking the AP exam, you can leave your answer at that point without having to simplify further. In class, though, follow whatever directions your instructor gives regarding how far you need to simplify.
Think of F(x) in terms of F(x) = 5x * (1 + x^2)^(-1) and you can use the product rule.
Product rule: If f(x) = a(x) * b(x), then f'(x) = a(x)*b'(x) + a'(x)*b(x)
In this case, a(x) = 5x and b(x) = (1 + x^2)^(-1)
a'(x) = 5
b'(x) = -1(1 + x^2)^(-2) * 2x by using the power rule). Simplifying, b'(x) = -2x/(1 + x^2)^2
So F'(x) = 5x * -2x/(1 + x^2)^2 + 5 * 1/(1 + x^2) = -10x^2 / (1 + x^2)^2 + 5 / (1 + x^2)
This could be simplified further, but that is not necessary since you just need to find F'(x) for a specific x = a. (In this case, a = 2.)
2. Evaluate F'(x) at x = 2.
F'(2) = -10 * 2^2 / (1 + 2^2)^2 + 5 / (1 + 2^2)
= -40 / 5^2 + 5 / 5
= -40/25 + 1
= -15/25
= -3/5
This gives you the slope of the line that is tangent to the curve F(x) at the point (2, 2).
3. Now use the slope of the tangent line m = -3/5 at the point (2,2) to find the equation of the tangent line.
y - y1 = m ( x - x1) where your slope is m and your point is (x1, y1)
If you are taking the AP exam, you can leave your answer at that point without having to simplify further. In class, though, follow whatever directions your instructor gives regarding how far you need to simplify.