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IB Mathematics AA SL Derivatives of xn , sinx , cosx , tanx Study Notes

IB Mathematics AA SL Derivatives of xn , sinx , cosx , tanx Study Notes - New Syllabus

IB Mathematics AA SL Derivatives of xn , sinx , cosx , tanx Study Notes

LEARNING OBJECTIVE

  • Derivative of xn (n ∈ ℚ), sinx, cosx, ex and lnx

Key Concepts: 

  • Basic Derivatives
  • Sum Rule & Constant Multiple Rule
  • Chain Rule
  • Product Rule & Quotient Rule
  • Application

MAA HL and SL Notes – All topics

Basic Derivatives

Basic Derivatives

The derivative of standard functions such as power functions, trigonometric, exponential, and logarithmic functions follows known formulas:

FunctionDerivative
\( f(x) = x^n \)   (where \( n \in \mathbb{Q} \))\( f'(x) = nx^{n-1} \)
\( f(x) = \sin x \)\( f'(x) = \cos x \)
\( f(x) = \cos x \)\( f'(x) = -\sin x \)
\( f(x) = e^x \)\( f'(x) = e^x \)
\( f(x) = \ln x \)\( f'(x) = \dfrac{1}{x} \)

Example

Differentiate the following functions:
a) \( f(x) = x^4 \)
b) \( f(x) = \sin x \)
c) \( f(x) = e^x \)
d) \( f(x) = \ln x \)

▶️ Answer/Explanation
Solution:
a) \( f'(x) = 4x^{3} \)
b) \( f'(x) = \cos x \)
c) \( f'(x) = e^x \)
d) \( f'(x) = \dfrac{1}{x} \)

Sum Rule & Constant Multiple Rule

Sum Rule & Constant Multiple Rule

Sum Rule:

\( \dfrac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x) \)

Constant Multiple Rule:

\( \dfrac{d}{dx}[a \cdot f(x)] = a \cdot f'(x) \), where \( a \) is a constant

Explanation:

You can differentiate term-by-term. Constants in front stay as multipliers

Example 

Differentiate the following:
a) \( f(x) = 3x^4 + 5x \)
b) \( f(x) = \sin x + \ln x \)
c) \( f(x) = 7e^x – 2x^3 \)

▶️ Answer/Explanation

Solution:

a) \( f'(x) = 3 \cdot 4x^3 + 5 = 12x^3 + 5 \)
b) \( f'(x) = \cos x + \dfrac{1}{x} \)
c) \( f'(x) = 7e^x – 2 \cdot 3x^2 = 7e^x – 6x^2 \)

Chain Rule (Composite Functions)

Chain Rule (Composite Functions)

If \( f(x) = h(g(x)) \), then

$f^{\prime}(x) = h{\prime}(g(x)) \cdot g{\prime}(x)$

In words:

Differentiate the outer function, keeping the inner function unchanged, then multiply by the derivative of the inner function.

Example 

Differentiate the following:
a) \( f(x) = \sin(3x) \)
b) \( f(x) = \ln(2x^2 + 1) \)
c) \( f(x) = e^{x^2 + 2} \)

▶️ Answer/Explanation

Solutions:

a) Let outer = \( \sin(u) \), inner = \( u = 3x \)
\( f'(x) = \cos(3x) \cdot 3 = 3\cos(3x) \)

b) Outer = \( \ln(u) \), inner = \( u = 2x^2 + 1 \)
\( f'(x) = \dfrac{1}{2x^2 + 1} \cdot 4x = \dfrac{4x}{2x^2 + 1} \)

c) Outer = \( e^u \), inner = \( u = x^2 + 2 \)
\( f'(x) = e^{x^2 + 2} \cdot 2x = 2x e^{x^2 + 2} \)

Product Rule & Quotient Rule

Product Rule & Quotient Rule

Product Rule:

$\frac{d}{dx}[f(x) \cdot g(x)] = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$

Quotient Rule:

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) – f(x)g'(x)}{(g(x))^2}$

Example:

Differentiate the following:
a) \( f(x) = x^2 \sin x \)
b) \( f(x) = \dfrac{e^x}{x^2} \)

▶️ Answer/Explanation

Solutions:

a) Use product rule: \( f'(x) = 2x \sin x + x^2 \cos x \)

b) Use quotient rule:
\( f'(x) = \dfrac{e^x \cdot x^2 – e^x \cdot 2x}{x^4} = \dfrac{e^x(x^2 – 2x)}{x^4} \)

Applications of Differentiation in Physics

Applications of Differentiation in Physics

Differentiation plays a critical role in describing motion and electromagnetic phenomena in physics. Two key examples are:

  • Uniform Circular Motion: Differentiation of position functions gives velocity and acceleration vectors.
  • Induced emf: Faraday’s law involves the rate of change of magnetic flux, modeled using derivatives.

 Example: Uniform Circular Motion

A particle moves in a circle of radius \( r = 5 \) m with angular velocity \( \omega = 2 \) rad/s. Its position is given by:
$ \vec{r}(t) = 5 \cos(2t)\, \hat{i} + 5 \sin(2t)\, \hat{j} $

  • a. Find the velocity vector \( \vec{v}(t) \)
  • b. Find the acceleration vector \( \vec{a}(t) \)
  • c. Show that the motion is uniform circular motion
▶️ Answer/Explanation

Step 1: Differentiate \( \vec{r}(t) \) to find velocity:

$ \vec{v}(t) = \frac{d\vec{r}}{dt} = -10 \sin(2t)\, \hat{i} + 10 \cos(2t)\, \hat{j} $

Step 2: Differentiate again to find acceleration:

$ \vec{a}(t) = \frac{d\vec{v}}{dt} = -20 \cos(2t)\, \hat{i} – 20 \sin(2t)\, \hat{j} $

Step 3: The acceleration points towards the center and has magnitude:

$ |\vec{a}(t)| = \sqrt{(-20 \cos(2t))^2 + (-20 \sin(2t))^2} = 20 $

This confirms the object is undergoing uniform circular motion with constant centripetal acceleration.

 Example: Induced emf using Faraday’s Law

A magnetic field through a coil changes over time according to:

$ \Phi(t) = 3t^2 – 2t \text{ (in Wb)} $ Use Faraday’s law to find the induced emf at time \( t = 2 \) s.

▶️ Answer/Explanation

Step 1: Use Faraday’s Law:

$ \mathcal{E} = -\frac{d\Phi}{dt} $ $ \frac{d\Phi}{dt} = \frac{d}{dt}(3t^2 – 2t) = 6t – 2 $

Step 2: Substitute \( t = 2 \):

$ \mathcal{E} = -(6(2) – 2) = -10 \text{ V} $

Conclusion: The induced emf at \( t = 2 \) s is −10 V.

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