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33 Cards in this Set
- Front
- Back
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The Product Rule
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[(fg)(x)'] = f(x)g(x)' + f'(x)g(x)
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The Quotient Rule
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[(f/g)(x)]' = [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2
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f '(x) of sin(x)
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cos(x)
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f '(x) of cos(x)
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-sin(x)
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f '(x) of tan(x)
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sec^2(x)
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f '(x) of cot(x)
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-csc^2(x)
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f '(x) of sec(x)
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sec(x)tan(x)
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f '(x) of csc(x)
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-cot(x)csc(x)
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The Chain Rule - Version 1
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Let "y" be y=F(u) where u = u(x).
Then y '(x) dy/dx = (dy/du) / (du/dx) |
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The Chain Rule - Version 2
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Let F(x) = f(g(x))
Then F '(x) = f ' (g(x))(g'(x)) |
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Chain Rule - Powers
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d/dx [u^r] = r (u)^r-1 (du/dx)
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Chain Rule of the Derivative of sin(u)
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cos(u)(du/dx)
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Chain Rule of the Derivative of cos(u)
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-sin(u)(du/dx)
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Chain Rule of the Derivative of tan(u)
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sec^2(u)(du/dx)
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Chain Rule of the Derivative of csc(u)
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-csc(u)cot(u)(du/dx)
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Chain Rule of the Derivative of sec(u)
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sec(u)tan(u)(du/dx)
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Chain Rule of the Derivative of cot(u)
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-csc^2(u)(du/dx)
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Steps for Implicit Differentiation
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1.) Treat "y" as a function of x
2.) Differentiate both sides of the equation, with respect to "x" and use the chain rule where appropriate. 3.) Solve algebraically for y' |
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f ' (x) [ln(x)]
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1/x
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f' (x) [e^x]
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e^x
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f ' (a^x)
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a^x(lna) where a is a constant
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Chain Rule for (e^u)'
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e^u(u')
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Chain Rule for (ln^u)'
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1/u (u)' = u'/u
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Chain Rule for (a^u)'
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a^u(lna)(u)'
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Chain Rule for (e^ax)'
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ae^ax
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ln (1)
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0
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ln (e)
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1
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ln (e^x)
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x, for any real number (x)
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e^lnx
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x
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ln(xy)
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ln(x) + ln(y) x>0, y>0
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ln (x/y)
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ln(x) - ln(y) x>0, y>0
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ln x^p
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p(ln(x)) where x>0 and p= any real number
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Steps for Logarithmic Differentiation
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1.) Replace f(x) with y
2.) take "ln" of both sides 3.) Use rules of logs to simplify step 2 4.) Differentiate both sides of step 3 with respect to x 5.) Solve for y ' |
