Chain Rule (Edexcel A-Level Mathematics): Revision Notes

If you have a function $y = f(g(x))$, where $g(x)$ is a function inside another function f, the derivative of $y$ with respect to $x$ is given by: $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

• $f'(g(x))$ is the derivative of the outer function evaluated at the inner function $g(x)$.
• $g'(x)$ is the derivative of the inner function $g(x)$ with respect to $x$.

In Leibniz notation, if y depends on u, and u depends on x (i.e., $y = f(u)\ and\ u = g(x)$), the chain rule can be written as: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ This shows that to find the derivative of y with respect to x, you multiply the derivative of y with respect to u by the derivative of u with respect to x.

To apply the chain rule effectively, follow these steps:

1. Identify the Outer and Inner Functions:

• Determine the outer function $f(u)$ and the inner function $u = g(x)$.

2. Differentiate the Outer Function:

• Differentiate $f(u)$ with respect to $u$, then substitute $g(x)$ back into the result.

3. Differentiate the Inner Function:

• Differentiate $g(x)$ with respect to $x$.

4. Multiply the Results:

• Multiply the derivative of the outer function by the derivative of the inner function.

Example 1: Differentiate $y = \sin(3x)$

• Step 1: Identify the functions:
• Outer function: $f(u) = \sin(u)$
• Inner function: $u = 3x$

• Step 2: Differentiate the outer function: $\frac{df}{du} = \cos(u)$

• Step 3: Differentiate the inner function: $\frac{du}{dx} = 3$

• Step 4: Apply the chain rule: $\frac{dy}{dx} = \cos(3x) \cdot 3 = 3\cos(3x)$

Example 2: Differentiate $y = \ln(5x^2 + 1)$

• Step 1: Identify the functions:
• Outer function: $f(u) = \ln(u)$
• Inner function: $u = 5x^2 + 1$

• Step 2: Differentiate the outer function: $\frac{df}{du} = \frac{1}{u}$

• Step 3: Differentiate the inner function: $\frac{du}{dx} = 10x$

• Step 4: Apply the chain rule: $\frac{dy}{dx} = \frac{1}{5x^2 + 1} \cdot 10x = \frac{10x}{5x^2 + 1}$

Example 3: Differentiate $y = e^{-\frac{1}{x}}$

• Step 1: Identify the functions:
• Outer function: $f(u) = e^u$
• Inner function: $u = -\frac{1}{x}$

• Step 2: Differentiate the outer function: $\frac{df}{du} = e^u$

• Step 3: Differentiate the inner function: $\frac{du}{dx} = \frac{1}{x^2}$

• Step 4: Apply the chain rule: $\frac{dy}{dx} = e^{-\frac{1}{x}} \cdot \frac{1}{x^2}$
• So,$\frac{dy}{dx} = \frac{e^{-\frac{1}{x}}}{x^2}.$

Example: Differentiate $y = \sin(\ln(3x))$

• Step 1: Identify the functions:
• Outermost function: $f(u) = \sin(u)$
• Middle function: $g(v) = \ln(v)$
• Innermost function: $h(x) = 3x$

• Step 2: Differentiate each function:
• $\frac{df}{du} = \cos(u)$
• $\frac{dg}{dv} = \frac{1}{v}$
• $\frac{dh}{dx} = 3$

• Step 3: Apply the chain rule: $\frac{dy}{dx} = \cos(\ln(3x)) \cdot \frac{1}{3x} \cdot 3 = \frac{\cos(\ln(3x))}{x}$

• The chain rule is essential for differentiating composite functions, allowing you to break down complex expressions into manageable parts.
• By identifying the inner and outer functions, differentiating each, and then multiplying, you can accurately compute derivatives for a wide range of functions.
• Mastery of the chain rule is crucial for solving advanced calculus problems and applications in various scientific and engineering disciplines.

Examples:

• $e^{x^2 + 2}$ ✔️
• $\ln(2x + 3)$ ✔️
• $x e^x$ ❌(Two separate functions multiplied together)

Example: If $y = e^{x^2 + 2}$, find $\frac{dy}{dx}$ 5. Let $u$ = "the most deeply nested part of the function," then write u = and then $y$ = in terms of $u$.

1. Find $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$.

6. Find $\dfrac{dy}{dx}$ using:

Fact:

Example: If $y = (x^4 + 3)^{10}$, find $\dfrac{dy}{dx}$ 7. Let $u = x^4 + 3$, then $y = u^{10}$.

Example: Find $f'(x)$ for $f(x) = \ln(e^x + x^2)$.

8. Let $u = e^x + x^2$.

9. $y = \ln(u)$.

10. Therefore,

Example: Find the first derivative of $y = \dfrac{1}{\sqrt{2x-3}}$.

11. Let $y = (2x-3)^{-\frac{1}{2}}$.

12. Let $u = 2x-3$.

13. $y = u^{-\frac{1}{2}}$.

14. Therefore,

Example: If $y = e^{x^2+2}$, find $\dfrac{dy}{dx}$.

15. Differentiate the "outer" expression as a whole.

16. Multiply by the inner expression differentiated.

Example: $y = (x^4 - 3)^10$, find $\dfrac{dy}{dx}$.

17. $y = (x^4 - 3)^{10}$

Example: Find $f'(x)$ for $f(x) = \ln(e^x + x^2)$.

18. $f(x) = \ln(e^x + x^2)$

Example: Find the equation of the tangent to $y = \sqrt{2x+1}$ when $x = 4$.

19. $y = (2x+1)^{\frac{1}{2}}$

20. Let $x = 4$:

21. Therefore,

22. Gradient = $\frac{1}{3}$, Point = $(4, 3)$.

23. Equation of the tangent:

Q6. (Jan 2010, Q5) The equation of a curve is $y = (x^2 + 1)^8$.

(i) Find an expression for $\frac{dy}{dx}$ and hence show that the only stationary point on the curve is the point for which $x = 0$.

Q2. (Jun 2006, Q1) Find the equation of the tangent to the curve $y = \sqrt{4x+1}$ at the point $(2, 3)$.

e.g. Justify that $y = \sqrt{2x+1}$ has no stationary points.

24. $y = (2x+1)^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{1}{\cancel2}(2x+1)^{-\frac{1}{2}} \times \cancel2 = (2x+1)^{-\frac{1}{2}} = 0$

1. Setting $\frac{dy}{dx} = 0$ leads to:

$\times (\sqrt{2x+1})\\ \leftarrow$ Intuitively, we only have control of the value of the denominator, and setting this to 0 would give an undefined answer, so no solutions.

25. This implies:

4. $\therefore \text{No solutions, so no stationary points.}$

e.g. Find $\frac{dy}{dx}$ when $y = x^2 + e^{x^2} + \ln(2x-3)$.

26. Split into parts:

• $\text{(A) } \frac{d}{dx}(x^2) = 2x \quad \text{(No chain rule needed)}$
• $\text{(B) } \frac{d}{dx}(e^{(x^2)}) = e^{(x^2)} \times 2x = 2xe^{x^2} \quad \text{(Chain rule is needed)}$
• $\text{(C) } \frac{d}{dx}(\ln(2x-3)) = \frac{1}{(2x-3)} \times 2 = \frac{2}{2x-3} \quad \text{(Chain rule is needed)}$

27. Combining all parts: