12892840
Description
Mind Map by ahmed hany, updated more than 1 year ago
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Created by GraceEChem
about 11 years ago
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Copied by ahmed hany
over 7 years ago
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Calculus
- Derivatives
- Product Rule
- (fg)' = f'g+g'f
- (fg)' = f'g+g'f
- Quotient Rule
- (f/g)' = (f'g - g'f) / g^2
- (f/g)' = (f'g - g'f) / g^2
- Power Rule for Functions
- (f^n)' = nf^(n-1)f'
- (f^n)' = nf^(n-1)f'
- Chain Rule
- dy/dx = (dy/du)(du/dx)
- for y=y(u) and u=u(x)
- for y=y(u) and u=u(x)
- [f(g(x)]' = f'(g(x))g'(x)
- dy/dx = (dy/du)(du/dx)
- Finding 2nd Derivative
- just take the
derivative of the
derivative
- just take the
derivative of the
derivative
- f(x) = e^x,
f'(x) = e^x
- f(x) = ln x,
f'(x) =1/x
- g(x) = loga(x),
g'(x) = (1/ln a)(1/x)
- f'(x) = ln(a)a^x
- f'(x) = ln(a)a^x
- Chain Rule
and Logs
- g(x) = ln(f(x))
- f'(x)/f(x)
- f'(x)/f(x)
- g(x) = ln(f(x))
- ln(x) + 1/x (x')
- f(x) = ln x,
f'(x) =1/x
- Product Rule
- Implicit Differentiation
- Treat x as x and
y as f(x), give
derivative of y
w/respect to x
- Differentiate both
sides w/respect to x
with y as f(x)
- Chain rule,
differentiating
terms with y
- Solve for dy/dx in
terms of x and y
- Solve for dy/dx in
terms of x and y
- Chain rule,
differentiating
terms with y
- Treat x as x and
y as f(x), give
derivative of y
w/respect to x
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