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
Find the equation of the tangent to the curve defined by
y= (lnx^2) / x
at the point where x=4.
5 answers
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.
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
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.
3 answers