Find an equation of the tangent line to the given curve at the specified point:
y=sqrt(x)/x+1, (4,2/5)
You can find the slope of the tangent line by setting the first derivative equal to zero and solving for x.
rewrite sqrt x as x^1/2
This becomes the derivative of a quotient.
The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator ALL OVER the denominator squared.
derivative of x^1/2 is (1/2)x^-1/2
derivative of x + 1 is 1
Can you put all of this together to find your slope?
Once you have the slope, find the equation of the line using the formula:
y = mx + b
you have x, m, y. Solve for b.
Then you will have the m and b that you need to substitute into y=mx + b
Is there anybody who can actually walk me through working out the problem please?
To find the equation of the tangent line to a curve at a specified point, we need to find the slope of the curve at that point, and then use the point-slope form of a linear equation.
1. First, find the derivative of the given function. The derivative will give us the slope of the curve at any point. In this case, the function is y = √(x)/(x+1).
To find the derivative of y with respect to x, we can use the quotient rule:
dy/dx = [(x+1)(1/2x^(-1/2)) - √x(1)]/(x+1)^2
Simplifying this expression gives:
dy/dx = (x+1)^(3/2) - √x / x(x+1)^2
2. Now substitute x = 4 into the derivative expression to find the slope at the point (4,2/5):
m = dy/dx |(x=4)
Substituting x = 4 into the derivative expression gives:
m = [(4+1)^(3/2) - √4] / 4(4+1)^2
Simplifying this expression gives:
m = [(5)^(3/2) - 2] / 100
3. Now we have the slope of the tangent line. Next, we will use the point-slope form of a linear equation, which is:
y - y₁ = m(x - x₁)
where (x₁, y₁) is the given point and m is the slope we just determined.
Substituting the values (x₁, y₁) = (4, 2/5) and m into the equation, we get:
y - 2/5 = [(5)^(3/2) - 2] / 100 (x - 4)
Simplifying this equation will give us the equation of the tangent line to the given curve at the specified point.