Find the equation of the tangent to the curve defined by
y= (lnx^2) / x
at the point where x=4.
To find the equation of the tangent to the curve, we first need to find the slope of the tangent at the point where x=4.
We can find the slope by taking the derivative of the function y= (lnx^2) / x and evaluating it at x=4.
y = (lnx^2) / x
y' = [(2/x) - ln(x^2)/x^2]
y'(4) = [(2/4) - ln(4^2)/4^2] = [0.25 - ln(16)/16] ≈ 0.1248
So the slope of the tangent at x=4 is approximately 0.1248.
Now we need to find the equation of the tangent, which can be done using the point-slope form of a line.
y - y1 = m(x - x1)
We know that the point (4, (ln4^2)/4) is on the tangent, and we just found that the slope at that point is approximately 0.1248.
Substituting these values into the point-slope form, we get:
y - ((ln4^2)/4) = 0.1248(x - 4)
Simplifying and rearranging, we get the equation of the tangent:
y = 0.1248x - 0.2951
no decimals, please use exact values (answer should include ln)
Okay, let's redo the solution using exact values.
The function is y = (ln(x^2))/x.
To find the slope of the tangent at x=4, we need to find y' and evaluate it at x=4.
y = (ln(x^2))/x
y' = (2/x - ln(x^2)/x^2)
y'(4) = (2/4 - ln(16)/16) = (1/2 - ln(2^4)/2^4) = (1/2 - ln(2)/16)
So the slope of the tangent at x=4 is (1/2 - ln(2)/16).
Now we use the point-slope form of a line to find the equation of the tangent. We know that the point (4, ln(16)/4) is on the tangent, and we just found that the slope is (1/2 - ln(2)/16).
The point-slope equation is:
y - y1 = m(x - x1)
Substituting in the values, we get:
y - ln(16)/4 = [(1/2 - ln(2)/16)](x - 4)
Simplifying and rearranging, we get the equation of the tangent:
y = (1/2)x - (ln(2)/4) - 1
Therefore, the equation of the tangent to the curve y = (ln(x^2))/x at the point x=4 is y = (1/2)x - (ln(2)/4) - 1.
y= (lnx^2) / x
y = 2lnx/x
dy/dx = ( x*2/x - 2lnx)/x^2
= (2 - 2lnx) / x^2
when x = 4, y = 2ln4/4 = ln4/2 , so the point of contact is (4, ln4/2)
dy/dx = (2 - 2ln4)/16 = (1-ln4)/8
equation : y - ln4/2 = (1-ln4)/8 * (x - 4)
I graphed mine and it works.
I graphed the bot's tangent equation and it's not even close to be
a tangent
I apologize for the error in my previous solution. Your solution and method are correct and I agree with your answer. Thank you for catching my mistake and providing the correct solution.
To find the equation of the tangent to the curve, we need to find both the slope and the y-intercept of the tangent line.
First, let's find the slope of the tangent line.
The slope of the tangent line can be found using the derivative of the curve. So, let's find the derivative of the curve y = (ln(x^2))/x.
Step 1: Rewrite the equation using the properties of logarithms:
y = ln(x^2) - ln(x)
Step 2: Simplify the equation:
y = 2ln(x) - ln(x)
= ln(x)
Step 3: Find the derivative of y with respect to x:
dy/dx = 1/x
Now, we have the slope of the tangent line at any point on the curve as dy/dx = 1/x.
Next, let's find the slope of the tangent line at x = 4.
We substitute x = 4 into the derivative:
dy/dx = 1/4
So, the slope of the tangent line at x = 4 is 1/4.
To find the y-intercept of the tangent line, we need to know the coordinate where the tangent line touches the curve. In this case, it is the point where x = 4.
Substitute x = 4 into the original equation y = (ln(x^2))/x:
y = (ln(4^2))/4
= (ln(16))/4
Now, we have the x-coordinate (4) and the y-coordinate ((ln(16))/4) of the point of tangency.
Finally, we have the slope (1/4) and the point (4, (ln(16))/4). We can use the point-slope form to find the equation of the tangent line:
y - y1 = m(x - x1)
Plugging in the values:
y - (ln(16))/4 = (1/4)(x - 4)
Simplifying the equation:
4y - ln(16) = x - 4
Rearranging the equation:
x - 4y = 4 + ln(16)
Thus, the equation of the tangent line to the curve at the point where x = 4 is x - 4y = 4 + ln(16).