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derivatives of trigonometric, exponential & logarithmic functions derivatives of trigonometric, exponential & logarithmic functions Derivatives Involving ln x MCV4U: Calculus & Vectors Recap Determine any value(s) of x where the tangent to f (x) = 2x · ln3 x is horizontal. Derivatives of General Exponential and Logarithmic Functions Use the product and chain rules to find the derivative. f 0 (x) = 2 ln3 x + 2x · 3 ln2 x · 1 x = 2 ln3 x + 6 ln2 x J. Garvin = 2 ln2 x (ln x + 3) J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 2/11 Slide 1/11 derivatives of trigonometric, exponential & logarithmic functions derivatives of trigonometric, exponential & logarithmic functions Derivatives Involving ln x Derivatives of Exponential Functions Since the tangent is horizontal, the slope is zero. Solve each factor. To find the derivative of a general exponential function, f (x) = b x where b 6= e, we can a technique similar to that used to find the derivative of f (x) = ln x. 2 ln2 x = 0 ln x + 3 = 0 2 y = bx ln x = −3 ln x = 0 ln y = ln b x x = e −3 ln x = 0 x =1 x= 1 e3 1 y Therefore, the tangent is horizontal when x = 1 or when x = e13 . · dy dx dy dx = ln b · x = ln b = y ln b = b x ln b Derivative of y = b x If f (x) = b x , then f 0 (x) = b x ln b. If y = b x , dy dx derivatives of trigonometric, exponential & logarithmic functions derivatives of trigonometric, exponential & logarithmic functions Derivatives of Exponential Functions Derivatives of Exponential Functions Example Example Determine the derivative of f (x) = 5x . Determine the slope of the tangent to y = This is pretty straightforward. dy dx Example Determine the derivative of f (x) = 3x 2 · 2x . Use the product rule. x 2 x f (x) = 6x · 2 + 3x · 2 ln 2 = 3x · 2x (2 + x ln 2) J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 5/11 2x when x = 4. x Use the quotient rule. f 0 (x) = 5x ln 5 0 = b x ln b. J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 4/11 J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 3/11 Thus, dy dx x=4 = 2x ln 2 · x − 2x x2 2x (x ln 2 − 1) = x2 = 24 (4 ln 2 − 1) = 4 ln 2 − 1. 42 J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 6/11 derivatives of trigonometric, exponential & logarithmic functions derivatives of trigonometric, exponential & logarithmic functions Derivatives of Logarithmic Functions Derivatives of Logarithmic Functions Using implicit differentiation, we can also determine the derivative of a general logarithmic function. Example Determine the derivative of f (x) = 6 log2 x + 5x. Recall that if y = logb x, then b y = x. Use the constant multiple and sum rules for derivatives. 6 f 0 (x) = +5 x ln 2 ln b y = ln x y ln b = ln x ln b · dy dx dy dx = 1 x = 1 x ln b Example Determine the slope of the tangent to y = log(3x + 2) when x = 1. Derivative of y = logb x If f (x) = logb x, then f 0 (x) = 1 x ln b . If y = logb x, dy dx = 1 x ln b . x=1 J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 8/11 J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 7/11 derivatives of trigonometric, exponential & logarithmic functions Derivatives of Logarithmic Functions derivatives of trigonometric, exponential & logarithmic functions Derivatives of Logarithmic Functions Example Example At what point on the function f (x) = log5 x is the slope The derivative is f 0 (x) = 1 x ln 5 . Set this equal to 1 25 1 x ln 5 25 ln 5 = x= f (x) = log5 Assuming log x is the common logarithm (base 10) of x, the 3 derivative is dy dx = (3x+2) ln 10 . Note the chain rule here. Therefore, dy = 5 ln3 10 . dx 25 ln 5 = 2 − log5 (ln 5). 1 25 ? 1 25 . 1 25 J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 9/11 derivatives of trigonometric, exponential & logarithmic functions J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 11/11 Using the chain rule, dy dx = 2 log3 x · dy Thus, dx = 2 log3 27 · 27 1ln 3 = x=27 Therefore, the function has a slope of at the point 25 ln 5 , 2 − log5 (ln 5) , or approximately (15.5, 1.7). Questions? 2 Determine the equation of the line parallel to y = (log 3 x) when x = 27, if it passes through the point 6, ln53 . 5 ln 3 5 ln 3 = = b= 1 x ln 3 . 2 9 ln 3 . 2 9 ln 3 (6) + 4 3 ln 3 + b 11 3 ln 3 The equation of the line is y = 2 9 ln 3 x J. Garvin — Derivatives of General Exponential andLogarithmic Functions Slide 10/11 b + 11 3 ln 3 .