Derivatives of Exponential, Logarithmic, and Hyperbolic Functions I. Quiz II. Homework III. The Derivative of f(x) = ex We will investigate the derivative of y = ex by sketching the graph. Then the students will complete the table:
Definition: e is the unique number such that if f(x) = ex then f'(x) = ex . Exercises: Find f'(x) if f(x) = A) 3ex B) xex C) ex /x D) e7x E) xe3ex Note that if f(x) = ex then f'(x) = lim h -> 0 [ex+h - ex ]/h. In particular if x = 0 then lim h-> 0 (eh -1)/h = 1 IV. Derivatives of Other Exponential Functions To find the derivative of f(x) = 2x we use the formula 2x = exln2 and the chain rule gives us f'(x) = ex ln 2 In particular if f(x) = bx then f'(x) = ex ln(b) V) The Derivative of the Natural Logarithm Function Recall that the natural logarithm function g(x) = ln(x) is the inverse of the exponential function f(x) = ex . To find f'(x) we use the inverse formula: g'(x) = 1/f'(g(x)) = 1/elnx = 1/x. Theorem: If f(x) = ln x then f'(x) = 1/x Example: Find f'(x) if f(x) = ex ln(x) Solution: We use the product rule f'(x) = ex ln x + ex /x VI. Hyperbolic Functions Exercises: Find f'(x) if A) cosh(x) = [ex + e-x]/2 B) sinh(x) = [ex - e-x]/2 C) tanh(x) = sinh(x)/cosh(x) D) coth(x) = cosh(x)/sinh(x) E) sech(x) = 1/cosh(x) F) csch(x) = 1/sinh(x) Proof that cosh2(x) - sinh2(x) = 1: Take derivatives of the left side: 2cosh(x)sinh(x) -2sinh(x)cosh(x) = 0 Hence the left hand side is a constant. Plugging in x = 1 gives cosh2(1) - sinh2(1) = 1.
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