91Ó°ÊÓ

Logarithmic functions have derivatives that can be quite interesting to work with. When you differentiate a logarithmic function like the natural log, specifically for an expression like \(\ln(u)\), the derivative is \(\frac{1}{u} \cdot \frac{du}{dx}\), where \(u\) is some function of \(x\). This rule comes from the fundamental definition of derivatives applied to logs. It's important because it shows how logs 'unwrap' functions, reducing the complexity you often face with exponentiated expressions. When handling these derivatives, remember to carefully identify and differentiate the 'inside' function \(u\), making the calculations simpler and more accurate as you'll see when applying the chain rule next.

The chain rule is your trusty tool for handling the derivative of composite functions. If you have a variable inside another function, the chain rule helps to properly differentiate it. For example, with \(y = \ln(x^2 + y^2)\), we apply the chain rule since our logarithmic expression includes \(x^2 + y^2\), rather than a simple \(x\).The chain rule states that if you have a function \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\). Here, for \(\ln(x^2 + y^2)\), \(f(u) = \ln(u)\) and \(g(x) = x^2 + y^2\). Applying the chain rule gives the derivative of the outside function \(\ln(u)\) as \(\frac{1}{u}\) then chain it with the derivative of the inside function \(g(x)\), which you'll extensively use along with other differentiation techniques.

Differentiation involves several techniques, one of which you've already seen with logarithms and the chain rule. In our example, we realize that differentiating \(y^2\) with respect to \(x\) isn't straightforward since \(y\) itself depends on \(x\).This situation requires implicit differentiation, where you assume \(y\) as a function of \(x\). By taking the derivative implicitly, you end up with terms that include \(y'\) since the original equation was in terms of both \(x\) and \(y\). It makes implicit differentiation incredibly useful when handling multi-variable equations. Using it, we came up with \(2yy'\) for the derivative of \(y^2\), showcasing how this method expands your differentiation toolkit and aids in finding solutions to such equations.

Once the differentiation process ends, the final touch is solving for \(y'\), the derivative itself. In our equation, after adding all the results from differentiation together, we arrive at an equation containing \(y'\) multiple times and other terms involving \(x\) and \(y\).The key is isolating \(y'\) on one side of the equation. You do this by collecting all the \(y'\) terms and factoring them out. For example, we organized our terms to get an expression like \(y'(x^2 + y^2 - 2y) = 2x\). Next, you divide both sides by the factor accompanying \(y'\) thus, solving for \(y'\). This method lets you cleanly and systematically extract \(y'\) so that the derivative is in its final, usable form: \(y' = \frac{2x}{x^2 + y^2 - 2y}\). When solving such expressions, double-check each algebraic step to ensure accuracy and to avoid common pitfalls.