Find the equation of the tangent line to the curve y = (9 ln x)/x at the points (1, 0) and (e, 9/e).
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Why did the tangent line to the curve go to therapy? Because it needed to find its equation! Let's get down to business and find out its equation step by step.
First, let's find the derivative of the given curve y = (9 ln x)/x. Using the quotient rule, the derivative is:
dy/dx = [(9/x)(1)] - [(9 ln(x))(1/x^2)]
= (9/x) - (9/x^2) ln(x)
Now, let's find the slope of the tangent line at the point (1, 0). Substituting x = 1 into the derivative:
m = (9/1) - (9/1^2) ln(1)
= 9 - 9(0)
= 9
So the slope of the tangent line at (1, 0) is 9. Now we can use the point-slope form of the equation of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the point (1, 0) and m is the slope we just found.
Plugging in the values, we have:
y - 0 = 9(x - 1)
Simplifying, we get:
y = 9x - 9
Now, let's find the slope of the tangent line at the point (e, 9/e). Substituting x = e into the derivative:
m = (9/e) - (9/e^2) ln(e)
= 9/e - 9/e^2
So the slope of the tangent line at (e, 9/e) is 9/e - 9/e^2. Now we can use the point-slope form again, but this time with the point (e, 9/e) and the slope 9/e - 9/e^2.
Putting it all together, the equation of the tangent line to the curve is:
y - (9/e) = (9/e - 9/e^2)(x - e)
Simplifying further, we have:
y = (9/e - 9/e^2)x + (9/e - 9/e^2)e - 9/e
Thus, the equation of the tangent line to the curve is:
y = (9/e - 9/e^2)x + 9 - 9e
To find the equation of the tangent line to the curve at a given point, you need to find the slope of the tangent line and the coordinates of the point of tangency. Once you have the slope and a point on the line, you can use the point-slope form to write the equation of the tangent line.
Given the curve y = (9 ln x) / x, to find the slope of the tangent line at any point (x, y) on the curve, we need to find the derivative of the curve with respect to x, which will give us the slope.
Let's start by finding the derivative of y with respect to x:
To simplify the expression, we'll apply the quotient rule.
The quotient rule states that for functions u(x) and v(x), the derivative of the quotient u(x) / v(x) is given by:
(d/dx)( u(x) / v(x) ) = (v(x) * du(x)/dx - u(x) * dv(x)/dx) / v(x)^2
In our case, u(x) = 9 ln x and v(x) = x. Applying the quotient rule:
dy/dx = (x * d/dx(9 ln x) - (9 ln x) * d/dx(x)) / (x^2)
To find d/dx(9 ln x), we need to use the chain rule. The derivative of ln x with respect to x is 1/x.
So, d/dx(9 ln x) = 9 * (1/x) = 9/x.
Also, d/dx(x) = 1.
Substituting these values back into the derivative expression:
dy/dx = (x * (9/x) - (9 ln x) * 1) / (x^2)
= (9 - 9 ln x) / (x^2)
Now that we have the derivative, we can find the slope of the tangent line at any point (x, y) on the curve by substituting the x-coordinate into the derivative expression.
To find the slope of the tangent line at (1, 0):
m = (9 - 9 ln 1) / (1^2)
= 9 - 9 * 0
= 9
So, the slope of the tangent line at (1, 0) is 9.
Similarly, to find the slope of the tangent line at (e, 9/e):
m = (9 - 9 ln e) / (e^2)
= 9 - 9 * 1
= 0
So, the slope of the tangent line at (e, 9/e) is 0.
Now we have the slope and a point on the tangent line, we can use the point-slope form to write the equation of the tangent line.
For the point (1, 0):
y - y1 = m(x - x1)
y - 0 = 9(x - 1)
y = 9x - 9
For the point (e, 9/e):
y - y1 = m(x - x1)
y - (9/e) = 0(x - e)
y = 9/e
Therefore, the equation of the tangent line to the curve y = (9 ln x) / x at the points (1, 0) and (e, 9/e) are:
- for the point (1, 0): y = 9x - 9
- for the point (e, 9/e): y = 9/e
y=0
y(x) = (9lnx)/x
y'(x) = 9(1-lnx)/x^2
y'(1) = 9
y'(e) = 0
Now just use the point-slope form of the lines.