Advanced Chain Rule Problems and Detailed Solutions

📚 Introduction to the Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions. A composite function is a function that is composed of another function, such as $f(g(x))$. The chain rule provides a way to find the derivative of such functions.

📜 History and Background

The chain rule was developed in the late 17th century by Gottfried Wilhelm Leibniz and Isaac Newton, independently. It emerged as a critical part of their work on calculus, providing a systematic way to handle complex derivatives that arise when one function is nested inside another. This discovery revolutionized mathematical analysis and laid the groundwork for many advanced techniques in calculus and related fields.

🔑 Key Principles

• 🔗 The Basic Idea: The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Mathematically, if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.
• 💡 Understanding the Layers: Think of the chain rule as peeling back layers of a function. You differentiate the outermost layer first, then move inward, differentiating each layer as you go, and multiplying the derivatives together.
• 📝 Notational Variations: The chain rule can also be expressed using Leibniz notation: if $y$ is a function of $u$, and $u$ is a function of $x$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

➗ Advanced Chain Rule Problems and Solutions

Let's dive into some more complex scenarios where the chain rule is essential. These problems often involve multiple layers of composition or require combining the chain rule with other differentiation rules.

1️⃣ Problem 1: Exponential Functions

Find the derivative of $y = e^{\sin(x^2)}$.

Solution:

• 🪜 Step 1: Identify the outer, middle, and inner functions. Here, we have $y = e^u$, where $u = \sin(v)$, and $v = x^2$.
• ⚙️ Step 2: Apply the chain rule. $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.
• ➗ Step 3: Calculate each derivative. $\frac{dy}{du} = e^u$, $\frac{du}{dv} = \cos(v)$, and $\frac{dv}{dx} = 2x$.
• 💡 Step 4: Substitute back to get $\frac{dy}{dx} = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x$.
• ✅ Final Answer: $\frac{dy}{dx} = 2x \cdot e^{\sin(x^2)} \cdot \cos(x^2)$.

2️⃣ Problem 2: Trigonometric Functions

Find the derivative of $y = \sin(\cos(\tan(x)))$.

Solution:

• 🪜 Step 1: Break down the composite functions. Let $y = \sin(u)$, where $u = \cos(v)$, and $v = \tan(x)$.
• ⚙️ Step 2: Apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.
• ➗ Step 3: Find the individual derivatives: $\frac{dy}{du} = \cos(u)$, $\frac{du}{dv} = -\sin(v)$, and $\frac{dv}{dx} = \sec^2(x)$.
• 💡 Step 4: Substitute back to express everything in terms of $x$: $\frac{dy}{dx} = \cos(\cos(\tan(x))) \cdot (-\sin(\tan(x))) \cdot \sec^2(x)$.
• ✅ Final Answer: $\frac{dy}{dx} = -\cos(\cos(\tan(x))) \cdot \sin(\tan(x)) \cdot \sec^2(x)$.

3️⃣ Problem 3: Logarithmic Functions

Find the derivative of $y = \ln(\sqrt{x^2 + 1})$.

Solution:

• 🪜 Step 1: Rewrite the function: $y = \ln((x^2 + 1)^{1/2}) = \frac{1}{2} \ln(x^2 + 1)$.
• ⚙️ Step 2: Apply the chain rule (or direct differentiation in this simplified form). Let $u = x^2 + 1$, so $y = \frac{1}{2} \ln(u)$.
• ➗ Step 3: Differentiate: $\frac{dy}{du} = \frac{1}{2u}$ and $\frac{du}{dx} = 2x$.
• 💡 Step 4: Apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2u} \cdot 2x = \frac{x}{x^2 + 1}$.
• ✅ Final Answer: $\frac{dy}{dx} = \frac{x}{x^2 + 1}$.

4️⃣ Problem 4: Power Rule and Chain Rule

Find the derivative of $y = (x^3 + 2x - 1)^4$.

Solution:

• 🪜 Step 1: Identify the outer and inner functions: $y = u^4$, where $u = x^3 + 2x - 1$.
• ⚙️ Step 2: Apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
• ➗ Step 3: Calculate derivatives: $\frac{dy}{du} = 4u^3$ and $\frac{du}{dx} = 3x^2 + 2$.
• 💡 Step 4: Substitute back to express everything in terms of $x$: $\frac{dy}{dx} = 4(x^3 + 2x - 1)^3 \cdot (3x^2 + 2)$.
• ✅ Final Answer: $\frac{dy}{dx} = 4(3x^2 + 2)(x^3 + 2x - 1)^3$.

5️⃣ Problem 5: Quotient and Chain Rule

Find the derivative of $y = \frac{\sin(2x)}{x^2 + 1}$.

Solution:

• 🪜 Step 1: Apply the quotient rule: $\frac{dy}{dx} = \frac{(x^2 + 1) \cdot (\sin(2x))' - \sin(2x) \cdot (x^2 + 1)'}{(x^2 + 1)^2}$.
• ⚙️ Step 2: Differentiate using the chain rule where needed. $(\sin(2x))' = 2\cos(2x)$ and $(x^2 + 1)' = 2x$.
• ➗ Step 3: Substitute the derivatives back into the quotient rule: $\frac{dy}{dx} = \frac{(x^2 + 1)(2\cos(2x)) - \sin(2x)(2x)}{(x^2 + 1)^2}$.
• 💡 Step 4: Simplify: $\frac{dy}{dx} = \frac{2(x^2 + 1)\cos(2x) - 2x\sin(2x)}{(x^2 + 1)^2}$.
• ✅ Final Answer: $\frac{dy}{dx} = \frac{2(x^2 + 1)\cos(2x) - 2x\sin(2x)}{(x^2 + 1)^2}$.

6️⃣ Problem 6: Inverse Trigonometric Functions

Find the derivative of $y = \arctan(e^{3x})$.

Solution:

• 🪜 Step 1: Recall that $\frac{d}{du}(\arctan(u)) = \frac{1}{1 + u^2}$.
• ⚙️ Step 2: Let $u = e^{3x}$. Then $y = \arctan(u)$.
• ➗ Step 3: Apply the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
• 💡 Step 4: Calculate derivatives: $\frac{dy}{du} = \frac{1}{1 + (e^{3x})^2} = \frac{1}{1 + e^{6x}}$ and $\frac{du}{dx} = 3e^{3x}$.
• ✅ Final Answer: $\frac{dy}{dx} = \frac{3e^{3x}}{1 + e^{6x}}$.

7️⃣ Problem 7: Multiple Nested Functions

Find the derivative of $y = \cos^3(\sin(2x^2))$.

Solution:

• 🪜 Step 1: Rewrite as $y = [\cos(\sin(2x^2))]^3$.
• ⚙️ Step 2: Break down the functions: $y = u^3$, where $u = \cos(v)$, $v = \sin(w)$, and $w = 2x^2$.
• ➗ Step 3: Apply the chain rule repeatedly: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}$.
• 💡 Step 4: Calculate the derivatives: $\frac{dy}{du} = 3u^2$, $\frac{du}{dv} = -\sin(v)$, $\frac{dv}{dw} = \cos(w)$, and $\frac{dw}{dx} = 4x$.
• ✅ Final Answer: $\frac{dy}{dx} = 3[\cos(\sin(2x^2))]^2 \cdot [-\sin(\sin(2x^2))] \cdot \cos(2x^2) \cdot 4x$.

📝 Conclusion

Mastering the chain rule involves understanding its underlying principles and practicing various types of problems. By breaking down composite functions and applying the chain rule methodically, you can successfully differentiate even the most complex functions. Keep practicing, and you'll become proficient in using this essential calculus tool!