Chương 4 || Đạo hàm và vi phân của hàm 1 biến || Quy tắc L' Hospital và Logarit hóa (Một số ví dụ)

Calculating Derivatives for Complex Functions
📌 The core topic involves finding the derivative of functions in the form $y = f(x)^{g(x)}$, which requires specialized techniques beyond basic differentiation rules.
📌 Method 1 (Logarithmic Differentiation): Take the natural logarithm ($\ln$) of both sides, use log properties to bring the exponent down (e.g., $\ln(y) = g(x) \ln(f(x))$), differentiate implicitly with respect to $x$, and then solve for $y'$.
📌 Method 2 (Using $e$): Rewrite the function as $y = e^{\ln(f(x)^{g(x)})} = e^{g(x) \ln(f(x))}$ and use the chain rule ($\frac{dy}{dx} = u' e^u$, where $u = g(x) \ln(f(x))$).
📌 For the example $y = x^{\cos(2x)}$, both methods yield the same result: $y' = x^{\cos(2x)} \left( -\sin(2x) \cdot 2 \ln(x) + \cos(2x) \cdot \frac{1}{x} \right)$.

L'Hôpital's Rule for Limits
📌 L'Hôpital's Rule is applicable when the limit results in an indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
📌 The rule states that $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
📌 Example 1: $\lim_{x\to 0} \frac{2^x - 1}{\sin(3x)}$ evaluates to $\frac{2^x \ln(2)}{3\cos(3x)} \to \frac{\ln(2)}{3}$ after applying the rule once.
📌 Example 2 involving trigonometric simplification: $\lim_{x\to 0} \frac{\tan(x) - \sin(x)}{x^3}$ is solved by applying L'Hôpital's rule multiple times or by algebraic manipulation resulting in $\frac{1}{2}$.

Logarithmic Differentiation for Indeterminate Forms in Limits
📌 For indeterminate forms like $1^\infty$ (e.g., $\lim_{x\to 0} (e^x + 3x)^{1/x}$), the technique of setting $y$ equal to the expression and taking the natural log is powerful.
📌 If $y = [f(x)]^{g(x)}$, then $\ln(y) = g(x) \ln(f(x))$. The problem reduces to finding the limit of $\ln(y)$ first.
📌 In the example $\lim_{x\to 0} (e^x + 3x)^{1/x}$, setting $y$ equal to the expression and taking the log resulted in finding $\lim_{x\to 0} \frac{\ln(e^x + 3x)}{x}$, which is the $\frac{0}{0}$ form.
📌 This limit of $\ln(y)$ was found to be $4$ using L'Hôpital's rule ($\frac{e^x + 3}{1} \to 1+3=4$). The final limit for $y$ is $e^4$.

Key Points & Insights
➡️ For derivatives of the form $f(x)^{g(x)}$, logarithmic differentiation is a mandatory technique when standard power or exponential rules fail.
➡️ L'Hôpital's Rule ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) provides a fast way to solve complex limits by repeatedly differentiating the numerator and denominator.
➡️ When dealing with limits resulting in the indeterminate form $1^\infty$, use the natural logarithm method to transform it into a solvable $\frac{0}{0}$ form suitable for L'Hôpital's Rule.

📸 Video summarized with SummaryTube.com on Nov 20, 2025, 16:44 UTC