Hello there! It's fantastic you're diving deeper into the Chain Rule – it's truly a cornerstone of calculus, especially when tackling more intricate functions. Don't worry, it's common for things to feel a bit tangled when combining rules, but with practice, it'll become second nature! Let's unravel some advanced problems together. 🤓
First, a quick refresher on the Chain Rule itself: If you have a composite function $f(g(x))$, its derivative is given by:
$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$
Or, if $y = f(u)$ and $u = g(x)$, then $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$This basically means "derivative of the outside function, keeping the inside intact, multiplied by the derivative of the inside function." Easy peasy, right? 😉
Example 1: Basic Nested Power
Problem: Find the derivative of $y = (3x^2 - 5x + 1)^4$.
Solution:
Here, the "outside" function is $(\text{stuff})^4$ and the "inside" is $3x^2 - 5x + 1$.
Let $u = 3x^2 - 5x + 1$. Then $y = u^4$.
$\frac{dy}{du} = 4u^3$
$\frac{du}{dx} = 6x - 5$
Applying the Chain Rule: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 4u^3 (6x - 5)$$
Substitute $u$ back in: $$\frac{dy}{dx} = 4(3x^2 - 5x + 1)^3 (6x - 5)$$
Example 2: Trigonometric & Nested
Problem: Differentiate $y = \cos(e^{2x})$.
Solution:
This is a double-nested function! The outermost is $\cos(\text{stuff})$, the next layer is $e^{\text{more stuff}}$, and the innermost is $2x$.
Step 1: Derivative of $\cos(\text{stuff})$ is $-\sin(\text{stuff})$.
$$\frac{d}{dx}[\cos(e^{2x})] = -\sin(e^{2x}) \cdot \frac{d}{dx}[e^{2x}]$$
Step 2: Now, we need the derivative of $e^{2x}$. Here, the outside is $e^{\text{stuff}}$ and the inside is $2x$.
$$\frac{d}{dx}[e^{2x}] = e^{2x} \cdot \frac{d}{dx}[2x] = e^{2x} \cdot 2 = 2e^{2x}$$
Combine them: $$\frac{dy}{dx} = -\sin(e^{2x}) \cdot (2e^{2x}) = -2e^{2x}\sin(e^{2x})$$
Example 3: Product Rule & Chain Rule Combined
Problem: Find the derivative of $y = x^3 \cdot \sqrt{5x+2}$.
Solution:
This function is a product $f(x)g(x)$ where $f(x) = x^3$ and $g(x) = \sqrt{5x+2} = (5x+2)^{1/2}$. We'll use the Product Rule: $$(fg)' = f'g + fg'$$
First, find $f'(x)$ and $g'(x)$.
$f'(x) = \frac{d}{dx}[x^3] = 3x^2$
For $g'(x)$, we need the Chain Rule: $g(x) = (5x+2)^{1/2}$.
Outside: $(\text{stuff})^{1/2}$. Inside: $5x+2$.
$g'(x) = \frac{1}{2}(5x+2)^{-1/2} \cdot \frac{d}{dx}[5x+2]$
$g'(x) = \frac{1}{2}(5x+2)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x+2}}$
Now, apply the Product Rule: $$\frac{dy}{dx} = (3x^2)(\sqrt{5x+2}) + (x^3)\left(\frac{5}{2\sqrt{5x+2}}\right)$$
To simplify further (optional but good practice):
$$\frac{dy}{dx} = 3x^2\sqrt{5x+2} + \frac{5x^3}{2\sqrt{5x+2}}$$
Find a common denominator: $$\frac{dy}{dx} = \frac{3x^2\sqrt{5x+2} \cdot 2\sqrt{5x+2} + 5x^3}{2\sqrt{5x+2}} = \frac{3x^2 \cdot 2(5x+2) + 5x^3}{2\sqrt{5x+2}}$$ $$\frac{dy}{dx} = \frac{6x^2(5x+2) + 5x^3}{2\sqrt{5x+2}} = \frac{30x^3 + 12x^2 + 5x^3}{2\sqrt{5x+2}} = \frac{35x^3 + 12x^2}{2\sqrt{5x+2}}$$
Your Turn! Practice Questions 🚀
Try these on your own. Solutions are hidden below!
- Question 1: Find $f'(x)$ if $f(x) = e^{\tan(x^2)}$.
- Question 2: Differentiate $y = \left(\frac{2x+1}{3x-2}\right)^3$.
- Question 3: Calculate $dy/dx$ for $y = x \sin(4x+1)$.
Solutions:
Question 1: $f'(x) = e^{\tan(x^2)} \cdot \sec^2(x^2) \cdot 2x = 2x \sec^2(x^2) e^{\tan(x^2)}$
Question 2: $$\frac{dy}{dx} = 3\left(\frac{2x+1}{3x-2}\right)^2 \cdot \frac{d}{dx}\left(\frac{2x+1}{3x-2}\right)$$
The derivative of the inside (using Quotient Rule):
$$ \frac{(2)(3x-2) - (2x+1)(3)}{(3x-2)^2} = \frac{6x-4-6x-3}{(3x-2)^2} = \frac{-7}{(3x-2)^2} $$
So, $$\frac{dy}{dx} = 3\left(\frac{2x+1}{3x-2}\right)^2 \cdot \frac{-7}{(3x-2)^2} = \frac{-21(2x+1)^2}{(3x-2)^4}$$
Question 3: Using Product Rule first, then Chain Rule for $\sin(4x+1)$:
$$ \frac{dy}{dx} = (1) \sin(4x+1) + x \cdot \cos(4x+1) \cdot 4 $$
$$ \frac{dy}{dx} = \sin(4x+1) + 4x \cos(4x+1) $$
Keep practicing these combinations, and you'll find that the Chain Rule is your best friend in differentiation! Remember to break down complex functions into layers. You've got this! 💪