#Derivatives of Composite Functions

#Table of Contents

1. Understanding Composite Functions
2. Using the Chain Rule
3. Worked Examples
4. Advanced Applications of the Chain Rule
5. Practice Questions
6. Glossary
7. Summary and Key Takeaways

#Understanding Composite Functions

When differentiating composite functions, we use the chain rule. Composite functions are of the form $f(g(x))$.

#The Chain Rule

If $y = f(u)$ and $u = g(x)$, then: $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

In function notation, if $h(x) = f(g(x))$, then: $$h'(x) = f'(g(x)) \cdot g'(x)$$

The chain rule can be extended to longer chains of functions: $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dr} \times \frac{dr}{dx}$$

The terms are derivatives, not fractions, but they can be treated similarly in this context to simplify the process.

#Using the Chain Rule

#Example: Differentiating $y = (2x^3 + 4x)^6$

1. Identify the inside function: $u = 2x^3 + 4x$
2. Rewrite the original function: $y = u^6$
3. Differentiate the inside function with respect to $x$: $$\frac{du}{dx} = 6x^2 + 4$$
4. Differentiate $y$ with respect to $u$: $$\frac{dy}{du} = 6u^5$$
5. Apply the chain rule: $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$ $$\frac{dy}{dx} = 6u^5 \times (6x^2 + 4)$$
6. Substitute back $u$: $$\frac{dy}{dx} = 6(2x^3 + 4x)^5 \times (6x^2 + 4)$$

To differentiate $y = (2x^3 + 4x)^6$: - Use the power as a coefficient and reduce the power by 1. - Multiply by the derivative of the inside function. - $$\frac{dy}{dx} = 6(2x^3 + 4x)^5 \times (6x^2 + 4)$$

#Worked Examples

#Example 1: $f(x) = \sin(3x^4 + 5x)$

1. Substitute the inside function: $$u = 3x^4 + 5x$$ $$y = \sin(u)$$
2. Differentiate $u$: $$\frac{du}{dx} = 12x^3 + 5$$
3. Differentiate $y$: $$\frac{dy}{du} = \cos(u)$$
4. Apply the chain rule: $$\frac{dy}{dx} = \cos(u) \cdot (12x^3 + 5)$$
5. Substitute back $u$: $$\frac{dy}{dx} = \cos(3x^4 + 5x) \cdot (12x^3 + 5)$$ $$f'(x) = (12x^3 + 5) \cos(3x^4 + 5x)$$

#Example 2: $g(x) = e^{\tan(x)}$

1. Substitute the inside function: $$u = \tan(x)$$ $$y = e^u$$
2. Differentiate $u$: $$\frac{du}{dx} = \sec^2(x)$$
3. Differentiate $y$: $$\frac{dy}{du} = e^u$$
4. Apply the chain rule: $$\frac{dy}{dx} = e^u \cdot \sec^2(x)$$
5. Substitute back $u$: $$\frac{dy}{dx} = e^{\tan(x)} \cdot \sec^2(x)$$ $$g'(x) = e^{\tan(x)} \sec^2(x)$$

#Advanced Applications of the Chain Rule

#Example: Combining Chain Rule with Product Rule

Differentiate $f(x) = \cos(x) \cdot e^{4x^3 + 3x^2 + 2}$:

1. Apply the product rule: Let $u = \cos(x)$ and $v = e^{4x^3 + 3x^2 + 2}$
2. Differentiate $u$: $$u' = -\sin(x)$$
3. Differentiate $v$ using the chain rule: Let $w = 4x^3 + 3x^2 + 2$, so $v = e^w$ $$\frac{dw}{dx} = 12x^2 + 6x$$ $$\frac{dv}{dw} = e^w$$ $$\frac{dv}{dx} = e^w \cdot (12x^2 + 6x)$$ $$v' = (12x^2 + 6x) e^{4x^3 + 3x^2 + 2}$$
4. Combine using the product rule: $$f'(x) = u'v + uv'$$ $$f'(x) = -\sin(x) \cdot e^{4x^3 + 3x^2 + 2} + \cos(x) \cdot (12x^2 + 6x) e^{4x^3 + 3x^2 + 2}$$
5. Factorize: $$f'(x) = e^{4x^3 + 3x^2 + 2} (\cos(x) (12x^2 + 6x) - \sin(x))$$

#Practice Questions

Practice Question

1. Differentiate $f(x) = \ln(\sin(2x^2 + 3))$.
2. Differentiate $g(x) = \sqrt{e^{x^2 + x}}$.
3. Differentiate $h(x) = \tan(\ln(x^2 + 1))$.
4. Differentiate $k(x) = e^{\cos(x^3 + x)}$.

#Glossary

• Chain Rule: A rule for differentiating a composite function.
• Composite Function: A function that is formed by combining two functions, denoted as $f(g(x))$.
• Derivative: A measure of how a function changes as its input changes.

#Summary and Key Takeaways

#Summary

• Composite functions are differentiated using the chain rule.
• The chain rule states that if $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
• This rule can be extended to longer chains of functions.
• When differentiating complex functions, the chain rule can be combined with other differentiation rules like the product or quotient rule.

#Key Takeaways

• Understand the inside function: Identify the inner function before applying the chain rule.
• Practice: The more you practice, the easier it becomes to apply the chain rule.
• Combine rules: Be ready to combine the chain rule with other differentiation rules for complex functions.

#Difficulty Rating

• Basic: Differentiating simple composite functions
• Intermediate: Combining the chain rule with the product or quotient rule
• Advanced: Differentiating nested composite functions

#Exam Strategy

By understanding and applying the chain rule effectively, you can tackle a wide range of differentiation problems with confidence. Happy studying!