Question

Tangent lines Find an equation of the line tangent to the graph of f at the given point.f(x) = sec−1(ex); (ln 2,π/3)

Accepted Answer

First, understand that the problem requires finding the equation of the tangent line to the function f(x) = sec^(-1)(e^x) at the point (ln 2, π/3). The general form of a tangent line equation is y - y1 = m(x - x1), where m is the slope of the tangent line and (x1, y1) is the given point. To find the slope m of the tangent line, we need to compute the derivative of f(x) = sec^(-1)(e^x). Recall that the derivative of sec^(-1)(u) with respect to x is 1 / (|u| * $$sqrt(u^2 - 1)) * du/dx. $$Here, u = e^x, so we need to apply the chain rule. Calculate the derivative of u = e^x, which is du/dx = e^x. Substitute this into the derivative formula for sec^(-1)(u) to get f'(x) = 1 / (|e^x| * sqrt((e^$$x)^2 - 1)) * e^x. $$Evaluate the derivative at the given point x = ln 2. Substitute x = ln 2 into f'(x) to find the slope m of the tangent line. This involves calculating e^(ln 2) and simplifying the expression. Finally, use the point-slope form of the line equation with the point (ln 2, π/3) and the slope m found in the previous step to write the equation of the tangent line. Substitute x1 = ln 2, y1 = π/3, and m into the equation y - y1 = m(x - x1).

Tangent lines Find an equation of the line tangent to the graph of f at the given point.
f(x) = sec⁻¹(e^x); (ln 2,π/3)

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Key Concepts

Tangent Line

Derivative

Inverse Functions