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

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

00:35

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Derivatives

Differentiation

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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Find an equation of the ta…

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Find an equation of the st…

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01:41

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.

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