math 1304 calculus i 3.5 and 3.6 – implicit and inverse functions
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Math 1304 Calculus I
3.5 and 3.6 – Implicit and Inverse Functions
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Implicit and Explicit Functions
• Explicit: y = f(x)
• Implicit: F(x,y)=0
Example:
2
22
1
1
xy
yx
implicit
explicit
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Implicit Differentiation
• If f(x) = g(x), then f’(x) = g’(x)
• Example: x2 + y2 = 1
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Inverse• f and g are inverse if:
y = f(x) iff x = g(y)• Also f and g are inverse if
f(g(y) = y and g(f(x) = x• Examples
Exponential and Logy = ln(x) iff x = ey
Trigonometric: sin and arcsiny = arcsin(x) iff x = sin(y)
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Derivatives of inverse functions
2
2
2
1
1)arctan(
1
1)arccos(
1
1)arcsin(
xx
dx
dx
xdx
dx
xdx
d
Proof? (in class)
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Derivative of Logarithms
• If F(x) = loga(f(x)), then F’(x) = (1/ln a) f’(x)/f(x)
• Proof? (in class)
• Special case:
If F(x) = ln(f(x)), then F’(x) = f’(x)/f(x)
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A new good working set of rules• Constants: If F(x) = c, then F’(x) = 0• Powers: If F(x) = f(x)n, then F’(x) = n f(x)n-1 f’(x), where n is real• Exponentials: If F(x) = af(x), then F’(x) = (ln a) af(x) f’(x)• Logarithms: If F(x) = loga(f(x)), then F’(x) = (1/ln a) f’(x)/f(x)• Trigonometric functions: If F(x) = sin(f(x)), then F’(x) = cos(f(x)) f’(x) If F(x) = csc(f(x)), then F’(x) = - csc(f(x)) cot(f(x)) f’(x) If F(x) = cos(f(x)), then F’(x) = - sin(f(x)) f’(x) If F(x) = sec(f(x)), then F’(x) = sec(f(x)) tan(f(x)) f’(x)
If F(x) = tan(f(x)), then F’(x) = sec2(f(x)) f’(x) If F(x) = cot(f(x)), then F’(x) = - csc2(f(x)) f’(x)• Inverse trig functions: If f(x) = arcsin(x), then f’(x) = 1/√ (1-x2) If f(x) = arccos(x), then f’(x) = -1/√ (1-x2) If f(x) = arctan(x), then f’(x) = 1/(1+x2)• Scalar multiplication: If F(x) = c f(x), then F’(x) = c f’(x)• Sum: If F(x) = g(x) + h(x), then F’(x) = g’(x) + h’(x)• Difference: If F(x) = g(x) - h(x), then F’(x) = g’(x) - h’(x)• Multiple sums: derivative of sum is sum of derivatives• Linear combinations: derivative of linear combination is linear combination of derivatives• Product: If F(x) = g(x) h(x), then F’(x) = g’(x) h(x) + g(x)h’(x)• Multiple products: If F(x) = g(x) h(x) k(x), then F’(x) = g’(x) h(x) k(x) + g(x) h’(x) k(x) + g(x) h(x) k’(x)• Quotient: If F(x) = g(x)/h(x), then F’(x) = (g’(x) h(x) - g(x)h’(x))/(h(x))2
• Composition: If F = fog is a composite, defined by F(x) = f(g(x)) then F'(x) = f'(g(x))g'(x)
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Logarithmic Differentiation
• Sometimes it helps to take the ln of both sides of an equation before differentiation.
• Then solve for y’
• Examples:
y = f(x)g(x)
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Use of logarithmic differentiation
• Prove general power law
• Quick proof of product rule