Implicit and Logarithmic Differentiation
Implicit Differentiation
- Definition: Implicit differentiation is a technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another.
- Example:
- Differentiate the implicit equation:
d(6 + (-)) = 4
dy/dx = y'
- General Rule for Implicit Differentiation: When differentiating an equation with respect to $x$, treat $y$ as a function of $x$ (i.e., use the chain rule).
Composite Functions
- Definition: A composite function is a function formed by applying one function to the results of another.
- Example:
- If $g(x)$ is the inner function and $f$ is the outer function, then $y = f(g(x))$.
- Differentiation of Composite Functions: Generally,
dy/dx = f'(g(x)) imes g'(x) - Relate this to the product of the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Derivatives of Logarithmic Functions
- Basic Rule:
- For the function $y = ext{log}_b{x}$, the derivative is given by
dy/dx = rac{1}{x imes ext{ln}(b)}
- Example of Differentiation:
- Differentiate the function $y = ext{log}_b{x}$:
dy/dx = rac{1}{x imes ext{ln}(b)} - In the case where $b=e$, the derivative simplifies to
dy/dx = rac{1}{x}
Logarithmic Rule
- Usage: Helpful for functions involving logarithms.
- Example: Differentiate the function $f(x) = x imes ext{ln}(x)$.
- Apply Product Rule:
f'(x) = x' imes ext{ln}(x) + x imes ( ext{ln}(x))'
= ext{ln}(x) + x imes rac{1}{x}
= ext{ln}(x) + 1
Differentiating Natural Logarithm
- Example Function:
f(x) = ext{ln}(rac{1}{x}) - Finding Domain (Df):
- The restrictions for the function include:
- |x| > 0
- Thus the domain is (- ext{∞}, 0) igcup (0, ext{∞})
- Derivative:
- For x > 0,
( ext{ln}(x))' = rac{1}{x} - For x < 0,
( ext{ln}(-x))' = rac{-1}{-x} = rac{1}{x}
Logarithmic Differentiation
- Process: Useful when differentiating complex functions.
- Example:
- Differentiate y = x^x.
- Transform: Take the natural logarithm on both sides:
ext{ln}(y) = ext{ln}(x^x)
ext{ln}(y) = x imes ext{ln}(x) - Differentiate implicitly with respect to x:
rac{d}{dx}( ext{ln}(y)) = rac{dy}{dx} imes rac{1}{y} = rac{d}{dx}(x imes ext{ln}(x)) - Apply product rule:
rac{dy}{dx} imes rac{1}{y} = (1 imes ext{ln}(x)) + (x imes rac{1}{x})
= ext{ln}(x) + 1 - Finally, re-arranging gives:
rac{dy}{dx} = y( ext{ln}(x) + 1).
Conclusion of Logarithmic Differentiation
- Implication: Logarithmic differentiation provides an efficient way to differentiate products/powers of variables that may prove difficult to handle otherwise.
- Key Takeaway: Use logarithmic differentiation when facing complex variable expressions, particularly when both bases and exponents involve the variable itself.