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

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SUMMARY

The equation of the tangent line to the function y=(x^2)lnx at the point (1,0) is derived using the first derivative. The first derivative is calculated as y'=2xlnx + x. Evaluating this at x=1 gives a slope of 1. Thus, the equation of the tangent line is y=x-1, confirming the solution is correct.

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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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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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