The image discusses the Chain Rule for differentiation, which applies when differentiating composite functions of the form $h(x) = f(g(x))$. The Chain Rule states that:

$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Breakdown of the Proof:

1. Starting with the definition of the derivative: $\frac{d}{dx}f(g(x)) = \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h}$ This is the standard difference quotient.

2. Introducing the change in $g(x)$:

   • By subtracting and adding $g(x+h) - g(x)$ within the numerator, the expression is rewritten as: $\lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{h} = \lim_{h \to 0} \frac{f(g(x+h)) - f(g(x))}{g(x+h) - g(x)} \cdot \frac{g(x+h) - g(x)}{h}$
   • This separates the change in $g(x)$ from the change in $f$, allowing the application of the Chain Rule.

3. Applying the limits:

   • Since $g(x)$ is differentiable, $\lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = g'(x)$.
   • The limit $\lim_{\Delta u \to 0} \frac{f(g(x) + \Delta u) - f(g(x))}{\Delta u} = f'(g(x))$.

4. Conclusion:

   • Putting everything together results in: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

Would you like more details or have any questions about this proof?

5 Related Questions:

1. How is the Chain Rule used in implicit differentiation?
2. What is the geometric interpretation of the Chain Rule?
3. How does the Chain Rule apply to higher-order derivatives?
4. Can you explain the difference between the Chain Rule and the Product Rule?
5. How can we apply the Chain Rule in cases where $g(x)$ is a trigonometric or exponential function?

Tip:

The Chain Rule can be extended to functions of multiple variables in multivariable calculus, which leads to the multivariable Chain Rule.