Let f (x) = ln[1 + g(x)] where g(7) = e4 − 1 and g′(7) = 6e4. Find the equation of the tangent line to f at x = 7.

Question

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Let f (x) = ln[1 + g(x)] where g(7) = e⁴ − 1 and g′(7) = 6e⁴. Find the equation of the tangent line to f at x = 7.

Answer 1

f(x) = ln[1 + g(x)]

f'(x) = [1/(1 + g(x))]g'(x)

Slope of tangent line = f'(7) = [1/ /(1 + g(7))]g'(7) = (1/e⁴)(6e⁴) = 6

Point on tangent line = (7,f(7)) = (7, ln(1+g(7)) = (7, 4)

Equation of tangent line: y - 4 = 6(x - 7)

So, y = 6x - 38