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# Find an equation of the tangent line to the curve at the given point.$y = xe^{-x^2}, (0, 0)$

## $y-0=1(x-0)$$y=x$

Derivatives

Differentiation

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### Video Transcript

Let's find the equation of the tangent line to this curve at the 0.0 The slope of the tangent line is going to be the derivative at the point. So let's find the derivative and our function is a product. So we need to use the product rule. So we have the first X times, the derivative of the second. And to find the derivative of the second, we have to use the chain rule. So we would get E to the negative X squared times, the derivative of negative X squared, which is negative two x So so far we have the first times, the derivative of the second, and we need plus the second times the derivative of the first and the derivative of X is just one. Okay, we could simplify this, but we could also just plug in r zero. Now we're evaluating this derivative at X equals zero. We could substitute the number in and then simplify later. So we have zero times e to the negative zero squared times negative to time zero. That's just all zero plus e to the negative zero squared times one Each of the zero is one So the derivative is one that's going to be the slope of the tangent line. So now, to find the equation of the line, we can use the point slope form. Why minus y one equals M times X minus X one, and we can substitute our 0.0 in for X one and why one? So we have y minus zero equals the slope one times X minus zero and that simplifies to B y equals X.

Oregon State University

Derivatives

Differentiation

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