Consider the function h(x)=e^x. Determine the equation for the tangent line at x = 1. All values must be exact (no decimal values).
can someone show me how you'd get to y=ex step by step??
To find the equation of the tangent line to the function h(x) = e^x at x = 1, we need to follow these steps:
Step 1: Find the derivative of the function h(x) = e^x.
Step 2: Evaluate the derivative at x = 1 to find the slope of the tangent line.
Step 3: Use the point-slope form of the equation of a line to determine the equation of the tangent line, using the slope from step 2 and the point (1, h(1)).
Step 1: Find the derivative of h(x) = e^x
The derivative of the function e^x can be found using the chain rule. Since the derivative of e^x is e^x itself, the derivative of h(x) = e^x is also e^x.
So, h'(x) = e^x.
Step 2: Evaluate the derivative at x = 1
To find the slope of the tangent line at x = 1, substitute x = 1 into the derivative h'(x) = e^x:
h'(1) = e^1 = e.
Step 3: Use the point-slope form to determine the equation of the tangent line
The point-slope form of the equation 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.
We already found the slope (m) in step 2 as e. For the point (x₁, y₁), we use x = 1 since we want the tangent line at x = 1.
So, x₁ = 1 and y₁ = h(1) = e^1 = e.
Now substitute these values into the point-slope form:
y - e = e(x - 1).
Simplifying, we get:
y - e = ex - e.
Adding e to both sides, we get:
y = ex.
Therefore, the equation of the tangent line to the function h(x) = e^x at x = 1 is y = ex.
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