Problem 59 - Calculus of a Single Variable

Question

Finding an Equation of a Tangent Line In Exercises \(55-62,\) find an equation of the tangent line to the graph of the function at the given point. $$ f(x)=e^{-x} \ln x, \quad(1,0) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of the function \(y=2^{-x}\) at the point \((-1,2)\) is \(y - 2 = -2\ln(2)(x -(-1))\) which equates to \(y = -2\ln(2)(x + 1) + 2\)

Step 1: Find the Derivative

Differentiate the function \(y = 2^{-x}\) with respect to x using the chain rule. The derivative of \(f(x) = a^{-x}\) (where 'a' is a constant) is \(-a^{-x} \ln(a)\). So the derivative \(y' = 2^{-x}\ln(2)\)

Step 2: Substituting the Given Point to Find the Slope of the Tangent Line

Plug in the given point \(-1, 2\) into the derivative to find the slope of the tangent line at that point. Replacing x by -1 in the derivative gives \(y' = 2^{-( -1)}\ln(2) = \(-2\ln(2)\)

Step 3: Generating the Equation of the Tangent Line

Use the point-slope form of the line equation \(y - y1 = m(x - x1)\), substituting -1 for x1, 2 for y1, and \(-2\ln(2)\) for m, we get the equation \(y - 2 = -2\ln(2)(x -(-1)) = -2\ln(2)(x + 1)\)

Key Concepts

Derivative Calculation

Understanding the derivative of a function is crucial when you're trying to find the equation of a tangent line. Derivative calculation involves finding the rate at which a function is changing at any point on its graph. For the function given in our exercise, \(y = 2^{-x}\), we calculate the derivative to find out how the function's output value changes as \(x\) changes. Put simply, if you imagine a curve on a graph, the derivative at any point is the slope of a line that just