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In Exercises \(55-62\) , find an equation of the tangent line to the graph of the function at the given point. $$y=e^{-2 x+x^{2}}, \quad(2,1)$$

Short Answer

Expert verified
The equation of the tangent line to the graph at the point (2,1) is \(y = 2x - 3\).

Step by step solution

01

Differentiate the given function

Apply the chain rule and product rule of differentiation: if \(u=e^{-2x}\) and \(v=e^{x^{2}}\), then \(y = u \cdot v\). According to the product rule, the derivative \(y'\) is \(u'v + uv'\). Calculate these derivatives: \(u' = -2e^{-2x}\) and \(v' = 2x \cdot e^{x^{2}}\). Substituting these into the product rule gives \(y' = -2e^{-2x} \cdot e^{x^{2}} + e^{-2x} \cdot 2x \cdot e^{x^{2}} = -2e^{x^{2}-2x} + 2x \cdot e^{x^{2}-2x} = (2x-2)e^{x^{2}-2x}\).
02

Find the slope of the tangent line

Evaluate this derivative at the given point (2,1) to find the slope of the tangent line at that point: \((2*2-2) e^{2^{2}-2*2} = 2e^{0} = 2\). So, the slope of the tangent line is 2.
03

Write the equation of the tangent line

Use the point-slope form of a linear equation, \(y - y_1 = m(x - x_1)\), where m is the slope and \((x_{1}, y_{1})\) are the coordinates of the point. Substituting the slope \(m = 2\) and the point \((x_{1}, y_{1}) = (2, 1)\) into this equation gives: \(y - 1 = 2(x - 2)\), which simplifies to \(y = 2x - 3\). This is the equation of the tangent line to the given function at the point (2,1).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
When facing a situation where you need to differentiate a function that is the product of two simpler functions, utilize the product rule. In essence, the product rule states that if you have two functions, let's call them u(x) and v(x), and you need to find the derivative of their product—u(x)v(x)—you don't just multiply the individual derivatives.

Instead, the derivative of the product, denoted as (u(x)v(x))', is u'(x)v(x) + u(x)v'(x). To remember this, think of saying 'first times the derivative of the second, plus the second times the derivative of the first'. In practice, if we label our functions u = e^{-2x} and v = e^{x^2}, then our derivative will involve the derivatives of both u and v, applied as the product rule indicates. This is vital for finding the slope of the tangent line to a curve at any given point.
Chain Rule
Whenever you have a function within a function, known as a composite function, the chain rule is your tool for differentiation. The chain rule can be a bit trickier to grasp, but it's incredibly powerful. Suppose you have a composite function f(g(x)). The chain rule tells us that to find the derivative of f(g(x)), denoted as f(g(x))', we need to take the derivative of the outer function f at g(x) and multiply it by the derivative of the inner function g(x).

This looks like f'(g(x)) * g'(x). For example, if you have e^(x^2), the exponent x^2 is the 'inner function,' and taking the 'e' to the power of something is the 'outer function.' The chain rule is essential for understanding how to differentiate more complex expressions and is used in conjunction with other rules like the product rule for functions like y = e^{-2x} * e^{x^2} addressed in our example.
Derivative
The derivative is one of the most significant concepts in calculus and provides the foundation for understanding how things change. It measures how a function's value changes as its input changes. The derivative of a function at a certain point gives us the slope of the tangent line to the function's graph at that point.

Mathematically, if y = f(x), the derivative of y with respect to x is denoted as f'(x) or y'. Calculating a derivative requires you to use rules like the product and chain rules, and a derivative tells us the rate at which y changes for a small change in x. In our exercise, differentiating y = e^{-2x} * e^{x^2} involves applying these rules to find the function that gives us the gradient of the tangent line at any point on the curve.
Slope of Tangent Line
The slope of a tangent line is a concrete measure of the steepness of the curve at a particular point. It's basically telling us how 'angled' the tangent is relative to the horizontal axis at that spot. For example, when we find the derivative of a function and evaluate it at a given point, what we are calculating is the slope of the tangent line at that point.

In our problem, after differentiating the function and finding the derivative with respect to x, we plug in the x-coordinate of the given point to find the slope, m. This tells us how sharply the function is curving at that point. A slope of 2 indicates that for every 1 unit increase in x, y increases by 2 units. The tangent line's equation uses this slope and the given point to provide a linear function that just grazes the curve at that spot, and it's an important tool for approximation and analysis in calculus.

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