Find equation tangent to y=(x^2)lnx at (1,0)

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In summary, the equation for the line y=(x^2)lnx at (1,0) is y=x-1. This was found by taking the first derivative of the equation, which gave y'=2xlnx +x. Using the formula for finding the slope at a given point, f'(1)=2(1)ln1 + 1 = 1, the slope of the function at the point (1,f(1)) was determined to be 1. Plugging this into the equation y=m(x-x_1) + y_1 and substituting the given point (1,0) for (x_1,y_1), the final equation y=x-1 was obtained.
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rmiller70015
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Homework Statement


Find an equation for the line y=(x^2)lnx at (1,0)


Homework Equations





The Attempt at a Solution


I took the first derivative of the equation and got y'=2xlnx +x
In other equations, it would be simple to find the slope, but at this point I am lost, is the slope 2? It doesn't appear to be in the form of y=mx+b.

So I've decided to use f'(1)=2(1)ln1 + 1 = 1 , this gives me the slope of the function at the point (1,f(1)) right?

If that's the case, then
y=m(x-x_1) + y_1
= 1(x-1) + 0
y=x-1

Just wanted to know if I am close or way off or maybe correct.
 
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  • #2
rmiller70015 said:

Homework Statement


Find an equation for the line y=(x^2)lnx at (1,0)


Homework Equations





The Attempt at a Solution


I took the first derivative of the equation and got y'=2xlnx +x
In other equations, it would be simple to find the slope, but at this point I am lost, is the slope 2? It doesn't appear to be in the form of y=mx+b.

So I've decided to use f'(1)=2(1)ln1 + 1 = 1 , this gives me the slope of the function at the point (1,f(1)) right?

If that's the case, then
y=m(x-x_1) + y_1
= 1(x-1) + 0
y=x-1

Just wanted to know if I am close or way off or maybe correct.

Looks ok to me.
 
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FAQ: Find equation tangent to y=(x^2)lnx at (1,0)

1. What is the equation of the tangent line to the curve y=(x^2)lnx at the point (1,0)?

The equation of a tangent line is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In order to find the equation of the tangent line, we first need to find the slope at the given point. This can be done by taking the derivative of the function y=(x^2)lnx and evaluating it at x=1. The derivative is given by y' = 2xlnx + x. Substituting x=1, we get the slope m = 2ln1 + 1 = 1. Therefore, the equation of the tangent line is y = x + b. To find the value of b, we plug in the coordinates of the given point (1,0). This gives us the equation 0 = 1 + b, which means b = -1. Thus, the equation of the tangent line is y = x - 1.

2. How do you determine if a point lies on a curve?

In order to determine if a point lies on a curve, we need to check if the coordinates of the point satisfy the equation of the curve. In this case, the point (1,0) lies on the curve y=(x^2)lnx if the equation y=(x^2)lnx holds true when x=1 and y=0. Plugging in the values, we get 0 = (1^2)ln1, which simplifies to 0=0. Therefore, the point (1,0) lies on the curve y=(x^2)lnx.

3. How do you find the slope of a curve at a given point?

The slope of a curve at a given point can be found by taking the derivative of the function representing the curve and evaluating it at the given point. In this case, the derivative of y=(x^2)lnx is y' = 2xlnx + x. Evaluating it at x=1, we get the slope m = 2ln1 + 1 = 1.

4. Can the equation of the tangent line be found at any point on the curve?

Yes, the equation of the tangent line can be found at any point on the curve. The process involves taking the derivative of the function representing the curve, evaluating it at the given point to find the slope, and then using the point-slope form of a line to find the equation of the tangent line.

5. How do you graph the tangent line at a given point?

To graph the tangent line at a given point, we can use the slope-intercept form of a line y=mx+b, where m is the slope and b is the y-intercept. First, we find the slope of the tangent line by taking the derivative of the function representing the curve and evaluating it at the given point. Then, we use the coordinates of the given point to find the value of b. Once we have the equation of the tangent line, we can plot the point and use the slope to find another point on the line. Finally, we can draw a straight line passing through these two points to graph the tangent line.

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