Find an equation of the tangent line to the graph of y = e-x2 at the point (2, 1/e4).
y' = -2x*e^(-x^2).
At x = 2, that the slope of the tangent line is:
y' = (-2)(2)e^[-(2)^2]
= -4e^(-4)
= -4/e^4.
tangent line is:
y - 1/e^4 = (-4/e^4)(x - 2)
y = (-4/e^4)x + 9/e^4.
Well, let's start by finding the derivative of the function y = e^(-x^2). That would be y' = -2x * e^(-x^2).
Now, let's find the slope of the tangent line at the point (2, 1/e^4). We substitute x = 2 into the derivative:
m = -2(2) * e^(-(2^2))
= -4 * e^(-4)
= -4/e^4
So, the slope of the tangent line is -4/e^4.
Now, let's use the point-slope form of the equation of a line: y - y1 = m(x - x1). Plugging in the values:
y - (1/e^4) = (-4/e^4)(x - 2)
Simplifying, we get:
y - 1/e^4 = -4x/e^4 + 8/e^4
And finally, rearranging the equation:
y = -4x/e^4 + (9/e^4)
And there you have it! The equation of the tangent line to the graph of y = e^(-x^2) at the point (2, 1/e^4) is y = -4x/e^4 + (9/e^4).
To find the equation of the tangent line to the graph of y = e^(-x^2) at the point (2, 1/e^4), we'll need to follow these steps:
Step 1: Find the derivative of the function y = e^(-x^2) using the chain rule.
Step 2: Evaluate the derivative at x = 2 to find the slope of the tangent line.
Step 3: Use the slope and the point (2, 1/e^4) to find the equation of the tangent line using the point-slope form.
Let's go through each step.
Step 1: Find the derivative of the function y = e^(-x^2).
To find the derivative, we need to use the chain rule. The derivative of e^u is e^u * u' (the derivative of the exponent times the derivative of the base function).
Taking the derivative of e^(-x^2) with respect to x, we get: dy/dx = (-2x) * e^(-x^2).
Step 2: Evaluate the derivative at x = 2.
Substituting x = 2 into the derivative, we get: dy/dx = (-2 * 2) * e^(-2^2) = -4 * e^(-4).
So, the slope of the tangent line at (2, 1/e^4) is -4 * e^(-4).
Step 3: Use the slope and the point (2, 1/e^4) to find the equation of the tangent line.
Using the point-slope form, which is y - y1 = m(x - x1), we can substitute the values: y - (1/e^4) = -4 * e^(-4) * (x - 2).
Expanding this equation, we have: y - (1/e^4) = -4 * e^(-4) * x + 8 * e^(-4).
Rearranging the equation, we get the final form: y = -4 * e^(-4) * x + 1/e^4 + 8 * e^(-4).
Therefore, the equation of the tangent line to the graph of y = e^(-x^2) at the point (2, 1/e^4) is y = -4 * e^(-4) * x + 1/e^4 + 8 * e^(-4).
To find the equation of the tangent line to the graph of y = e^(-x^2) at the point (2, 1/e^4), we need to find the slope of the tangent line at that point.
The slope of a tangent line is given by the derivative of the function. So, we'll start by finding the derivative of y = e^(-x^2).
Step 1: Find the derivative of y = e^(-x^2)
Let's use the chain rule to find the derivative:
dy/dx = d/dx(e^(-x^2))
To differentiate e^(-x^2), we need to take the derivative of the exponential function and the exponent.
The derivative of e^u with respect to u is e^u, and the derivative of -x^2 with respect to x is -2x.
Applying the chain rule, we have:
dy/dx = e^(-x^2) * (-2x)
Now, we have the slope of the tangent line at any point (x, y).
Step 2: Find the slope at the point (2, 1/e^4)
To find the slope at (2, 1/e^4), substitute x = 2 into dy/dx:
m = dy/dx |(x=2)
= e^(-2^2) * (-2 * 2)
= e^(-4) * (-4)
= -4e^(-4)
Now, we have the slope of the tangent line.
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is given by:
y − y₁ = m(x − x₁)
where (x₁, y₁) is a point on the line, and m is the slope.
Using the point (2, 1/e^4) and slope -4e^(-4), we can write the equation of the tangent line:
y − 1/e^4 = -4e^(-4)(x − 2)
Finally, simplify the equation if desired:
y − 1/e^4 = -4e^(-4)x + 8e^(-4)
y = -4e^(-4)x + 8e^(-4) + 1/e^4
y = -4e^(-4)x + 9/e^4
So, the equation of the tangent line to the graph of y = e^(-x^2) at the point (2, 1/e^4) is y = -4e^(-4)x + 9/e^4.