notes_logs_and_implicit_diff

# Dy 220 01 0 04 therefore at x 2 if x is increased by

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dy = 2(2)(0 . 01) = 0 . 04 Therefore, at x = 2 , if x is increased by 0.01 then y will increase by 0.04. 1.3 Implicit Di ff erentiation Suppose we have the following: 2 y + 3 x = 12 we can rewrite it as 2 y = 12 3 x y = 6 3 2 x Now we have y = f ( x ) and we can take the derivative dy dx = 3 2 Lets consider an alternative. We know that y is a function of x or, y = y ( x ) and the derivative of y is dy dx . If we return to our original equation, 2 y + 3 x = 12 , we can di ff erentiate it IMPLICITLY in the following manner 2 y + 3 x = 12 2 dy + 3 dx = 0 μ d (12) dx = 0 2 dy dx + 3 dx dx = 0 2 dy dx + 3 = 0 μ dx dx = 1 2

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rearrange to get dy dx by itself 2 dy dx = 3 dy dx = 3 2 which is what we got before! Here is a few more examples: 1. y 2 + x 2 = 36 2 ydy + 2 xdx = d (36) 2 y dy dx + 2 x dx dx = 0 μ remember d (36) dx = 0 2 y dy dx + 2 x = 0 dy dx = 2 x 2 y = x y 2. 5 y 3 + 4 x 5 = 250 15 y 2 dy dx + 20 x 4 = 0 15 y 2 dy dx + 20 x 4 = 0 dy dx = 20 x 4 15 y 2 = 4 x 4 3 y 2 3. y 1 / 2 2 x 2 + 5 y = 15 1 2 y 1 / 2 dy 4 xdx + 5 dy = 0 μ 1 2 y 1 / 2 + 5 dy dx 4 x = 0 ( ÷ by dx ) dy dx = 4 x ¡ 1 2 y 1 / 2 + 5 ¢ When you are using implicit di ff erentiation, there are two things to remember: First: All the rules apply as before Second: you are ASSUMING that you can rewrite the equation in the form y = f ( x ) Example: Special application of the product rule. Suppose you want to implicitly di ff erentiate xy = 24 what do we do here?
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