Understanding Implicit Differentiation: A Detailed Guide
Implicit differentiation is a powerful technique in calculus used to find the derivative of an implicit function. These functions are often given without being solved explicitly for one variable in terms of the other. In this guide, we will delve into the process of implicit differentiation through a detailed example, making the concept more accessible and understandable.
Introduction to Implicit Differentiation
Implicit functions are equations where the dependent variable is not isolated and is expressed in terms of both the independent and dependent variables. For example, (x^y y^x 1). In such cases, the derivative (frac{dy}{dx}) must be found without explicitly solving for (y).
Step-by-Step Example: Differentiating xy yx1
Let's consider the function (x^y y^x 1). To differentiate this with respect to (x), we will use the sum rule and implicit differentiation.
Step 1: Apply the Sum Rule
The sum rule states that the derivative of a sum is the sum of the derivatives. Therefore:
[frac{d}{dx}(x^y y^x) frac{d}{dx}(1)]
Since the derivative of a constant (1) is 0, we have:
[frac{d}{dx}x^y frac{d}{dx}y^x 0]
Step 2: Differentiate xy
To differentiate (x^y) with respect to (x), we use the power rule and the chain rule:
[frac{d}{dx}x^y yx^{y-1} cdot ln(x) x^y cdot frac{dy}{dx}]
[text{(first term for } x text{ differentiation, second term for } y text{ differentiation)}
Step 3: Differentiate yx
Similarly, to differentiate (y^x) with respect to (x), we use the power rule and the chain rule:
[frac{d}{dx}y^x x y^{x-1} cdot ln(y) y^x cdot frac{dy}{dx}]
[text{(first term for } y text{ differentiation, second term for } y text{ differentiation)}
Step 4: Combine the Derivatives
Now, we combine the two derivatives:
[left(yx^{y-1} cdot ln(x) x^y cdot frac{dy}{dx}right) left(x y^{x-1} cdot ln(y) y^x cdot frac{dy}{dx}right) 0]
Step 5: Isolate dydx
Group terms involving (frac{dy}{dx}) on one side:
[frac{dy}{dx} left( x^y cdot ln(x) y^x cdot ln(y) right) - left( yx^{y-1} cdot ln(x) x y^{x-1} cdot ln(y) right)]
Divide both sides by the coefficient of (frac{dy}{dx}):
[frac{dy}{dx} - frac{yx^{y-1} cdot ln(x) x y^{x-1} cdot ln(y)}{x^y cdot ln(x) y^x cdot ln(y)}]
Step 6: Simplify the Expression (Optional)
Further simplification can be done, but it's not always necessary:
[frac{dy}{dx} - frac{yx^{y-1} cdot ln(x) x y^{x-1} cdot ln(y)}{x^y cdot ln(x) y^x cdot ln(y)}]
This can be simplified by factoring common terms, but it already provides a clear and simplified answer:
[frac{dy}{dx} - frac{y cdot frac{x^y}{x} x cdot frac{y^x}{y} cdot ln(y)}{x^y cdot ln(x) y^x cdot ln(y)}]
The expression can be further simplified by multiplying the numerator and denominator by (xy), but it is already a functional representation.
Conclusion
This detailed walkthrough should help you understand how to apply implicit differentiation to complex functions. The key is to isolate (frac{dy}{dx}) and use the chain rule and power rule effectively.