All Study Guides Differential Calculus Unit 9
∬ Differential Calculus Unit 9 – Inverse Function & Logarithm DerivativesInverse functions and logarithm derivatives are essential tools in calculus. They allow us to reverse mathematical operations and simplify complex calculations. These concepts are crucial for solving equations, modeling real-world phenomena, and analyzing rates of change.
Mastering inverse functions and logarithm derivatives opens doors to advanced mathematical techniques. These skills are fundamental in fields like physics, engineering, and economics, where they're used to solve complex problems and make predictions about dynamic systems.
Key Concepts
Inverse functions reverse the input and output of the original function
Logarithmic functions are the inverse of exponential functions
log b ( x ) \log_b(x) log b ( x ) is the power to which b b b must be raised to get x x x
The derivative of an inverse function is related to the derivative of the original function
Involves reciprocals and the chain rule
Logarithm derivatives utilize the properties of logarithms and the chain rule
Logarithmic differentiation is useful for products, quotients, and powers
Applications include solving equations, modeling growth, and analyzing rates of change
Common mistakes involve incorrect use of properties or forgetting to apply the chain rule
Practice problems reinforce understanding and develop problem-solving skills
Inverse Functions Basics
An inverse function, denoted as f − 1 ( x ) f^{-1}(x) f − 1 ( x ) , "undoes" the original function f ( x ) f(x) f ( x )
If f ( a ) = b f(a) = b f ( a ) = b , then f − 1 ( b ) = a f^{-1}(b) = a f − 1 ( b ) = a
To find the inverse function, swap x x x and y y y in the original function and solve for y y y
Example: If f ( x ) = 2 x + 1 f(x) = 2x + 1 f ( x ) = 2 x + 1 , then f − 1 ( x ) = x − 1 2 f^{-1}(x) = \frac{x-1}{2} f − 1 ( x ) = 2 x − 1
The graphs of a function and its inverse are reflections across the line y = x y = x y = x
A function must be one-to-one (injective) to have an inverse
Horizontal line test: If any horizontal line intersects the graph more than once, the function is not one-to-one
The domain of a function becomes the range of its inverse, and vice versa
Composition of a function with its inverse results in the identity function
f ( f − 1 ( x ) ) = f − 1 ( f ( x ) ) = x f(f^{-1}(x)) = f^{-1}(f(x)) = x f ( f − 1 ( x )) = f − 1 ( f ( x )) = x
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions
If y = b x y = b^x y = b x , then x = log b ( y ) x = \log_b(y) x = log b ( y ) , where b b b is the base
Common logarithmic bases include 10 (common log), e e e (natural log), and 2 (binary log)
Properties of logarithms:
log b ( x y ) = log b ( x ) + log b ( y ) \log_b(xy) = \log_b(x) + \log_b(y) log b ( x y ) = log b ( x ) + log b ( y )
log b ( x y ) = log b ( x ) − log b ( y ) \log_b(\frac{x}{y}) = \log_b(x) - \log_b(y) log b ( y x ) = log b ( x ) − log b ( y )
log b ( x n ) = n log b ( x ) \log_b(x^n) = n\log_b(x) log b ( x n ) = n log b ( x )
log b ( 1 ) = 0 \log_b(1) = 0 log b ( 1 ) = 0 and log b ( b ) = 1 \log_b(b) = 1 log b ( b ) = 1
Change of base formula: log b ( x ) = log a ( x ) log a ( b ) \log_b(x) = \frac{\log_a(x)}{\log_a(b)} log b ( x ) = l o g a ( b ) l o g a ( x )
Logarithmic functions are continuous and increasing for bases greater than 1
The graph of a logarithmic function has a vertical asymptote at x = 0 x = 0 x = 0 and grows slowly for large x x x
Derivative Rules for Inverse Functions
The derivative of an inverse function is related to the derivative of the original function
If y = f − 1 ( x ) y = f^{-1}(x) y = f − 1 ( x ) , then d y d x = 1 f ′ ( y ) \frac{dy}{dx} = \frac{1}{f'(y)} d x d y = f ′ ( y ) 1
To find the derivative of an inverse function:
Replace x x x with y y y in the original function
Differentiate the original function with respect to y y y
Solve for d y d x \frac{dy}{dx} d x d y by taking the reciprocal
Replace y y y with f − 1 ( x ) f^{-1}(x) f − 1 ( x )
The chain rule is often necessary when applying the inverse function derivative rule
Example: If y = ( 2 x + 1 ) 3 y = (2x + 1)^3 y = ( 2 x + 1 ) 3 , then d y d x = 1 6 ( 2 y 1 / 3 − 1 ) \frac{dy}{dx} = \frac{1}{6(2y^{1/3} - 1)} d x d y = 6 ( 2 y 1/3 − 1 ) 1
Inverse trigonometric functions (arcsin, arccos, arctan) have specific derivative rules
Example: d d x arcsin ( x ) = 1 1 − x 2 \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}} d x d arcsin ( x ) = 1 − x 2 1
Logarithm Derivative Techniques
Logarithmic differentiation is a technique for finding derivatives of complicated functions
Involves taking the natural log of both sides of an equation and using properties of logarithms
Useful for differentiating products, quotients, and functions raised to variable powers
Steps for logarithmic differentiation:
Take the natural log of both sides of the equation
Use logarithm properties to simplify the right-hand side
Differentiate both sides with respect to x x x using the chain rule
Solve for d y d x \frac{dy}{dx} d x d y by multiplying both sides by y y y
Example: To find d d x ( x x ) \frac{d}{dx} (x^x) d x d ( x x ) :
ln ( y ) = ln ( x x ) = x ln ( x ) \ln(y) = \ln(x^x) = x\ln(x) ln ( y ) = ln ( x x ) = x ln ( x )
1 y d y d x = ln ( x ) + 1 \frac{1}{y} \frac{dy}{dx} = \ln(x) + 1 y 1 d x d y = ln ( x ) + 1
d y d x = x x ( ln ( x ) + 1 ) \frac{dy}{dx} = x^x(\ln(x) + 1) d x d y = x x ( ln ( x ) + 1 )
Logarithmic differentiation is also helpful for differentiating functions with multiple factors
Example: y = ( 2 x + 1 ) 3 ( 3 x − 2 ) 2 y = (2x + 1)^3 (3x - 2)^2 y = ( 2 x + 1 ) 3 ( 3 x − 2 ) 2
Applications and Examples
Inverse functions are used to solve equations and model real-world situations
Example: If a car's value depreciates according to V ( t ) = 20000 ( 0.8 ) t V(t) = 20000(0.8)^t V ( t ) = 20000 ( 0.8 ) t , find when the value will be $5000
Logarithmic functions model exponential growth and decay
Example: Population growth, radioactive decay, compound interest
Logarithmic differentiation is applied in physics, economics, and engineering
Example: Analyzing the elasticity of demand in economics
Inverse trigonometric functions appear in calculus, physics, and geometry
Example: Calculating the angle of elevation in a right triangle
Derivatives of inverse functions and logarithms are used in optimization problems
Example: Minimizing the cost of production in a manufacturing process
Understanding these concepts is crucial for advanced mathematics and science courses
Example: Differential equations, complex analysis, and mathematical modeling
Common Mistakes and Tips
Forgetting to use the chain rule when differentiating inverse or logarithmic functions
Tip: Identify the "inner" and "outer" functions and apply the chain rule
Incorrectly applying logarithm properties or confusing ln and log
Tip: Practice using properties and remember ln is base e, log is base 10
Misinterpreting the domain and range of inverse functions
Tip: Sketch the graph of the original function to identify the domain and range
Neglecting to simplify expressions or cancel terms when possible
Tip: Look for common factors or logarithms that can be combined
Making algebraic errors when solving for the inverse function or derivative
Tip: Double-check your work and use online resources to verify your answers
Not checking the conditions for the existence of an inverse function
Tip: Use the horizontal line test to ensure the function is one-to-one
Overcomplicating the problem by using unnecessary techniques
Tip: Identify the most efficient method for the given problem
Practice Problems
Find the inverse of the function f ( x ) = 3 x − 2 x + 1 f(x) = \frac{3x - 2}{x + 1} f ( x ) = x + 1 3 x − 2
Determine the derivative of y = log 5 ( 2 x + 1 ) y = \log_5(2x + 1) y = log 5 ( 2 x + 1 )
Use logarithmic differentiation to find d d x ( x 2 + 1 ) 3 \frac{d}{dx} (x^2 + 1)^3 d x d ( x 2 + 1 ) 3
Calculate the derivative of y = arccos ( 3 x − 1 ) y = \arccos(3x - 1) y = arccos ( 3 x − 1 )
If f ( x ) = e 2 x f(x) = e^{2x} f ( x ) = e 2 x and g ( x ) = ln ( x + 1 ) g(x) = \ln(x + 1) g ( x ) = ln ( x + 1 ) , find ( f ∘ g ) ′ ( x ) (f \circ g)'(x) ( f ∘ g ) ′ ( x )
Evaluate lim x → 0 ln ( 1 + 2 x ) x \lim_{x \to 0} \frac{\ln(1 + 2x)}{x} lim x → 0 x l n ( 1 + 2 x )
Prove that d d x arctan ( x ) = 1 1 + x 2 \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2} d x d arctan ( x ) = 1 + x 2 1
Find the equation of the tangent line to the curve y = x e − x y = xe^{-x} y = x e − x at the point ( 1 , 1 e ) (1, \frac{1}{e}) ( 1 , e 1 )
Determine the value of x x x for which the function y = x log 2 ( x ) y = x\log_2(x) y = x log 2 ( x ) has a horizontal tangent
Solve the equation e 2 x − 3 e x + 2 = 0 e^{2x} - 3e^x + 2 = 0 e 2 x − 3 e x + 2 = 0 using the substitution u = e x u = e^x u = e x