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AS Level
Mathematics 9709 P1
7. Differentiation
Written by: Tharun Athreya
Formatted by: Dhyaneshwaran V
Index
7.1 Derivatives and gradient functions
- The process of calculating gradients of a curve at any point is called differentiation.
- At any point on the curve, the gradient of the curve is equal to the gradient of the tangent at that point.
- A tangent should touch only one point on the curve.
- General Formula:
\(\frac{d}{dx} (x^n) = n x^{n-1}\) - There are different ways to represent the first derivative:
- If \( y = x^3 \), then \( \frac{dy}{dx} = 3x^2 \)
- If \( f(x) = x^3 \), then \( f'(x) = 3x^2 \)
- \(\frac{d}{dx} (x^3) = 3x^2\)
- Remember to multiply the power by \( n \) and then subtract one from the power.
- You need to know two rules and be able to use them:
- Scalar Multiple Rule:
\(\frac{d}{dx} [k f(x)] = k \frac{d}{dx} [ f(x) ]\), where \( k \) is a constant. This rule applies when there is one function and a constant. - Addition Rule:
\(\frac{d}{dx} [ f(x) + g(x)] = \frac{d}{dx} [f(x)] + \frac{d}{dx} [g(x)]\) This rule also applies to subtraction; just replace the addition sign with a subtraction sign.
Example:
- For \( y = 3x^2 \), \(\frac{dy}{dx} = 6x\).
- For \( y = 5x^3 + 2x \), \(\frac{dy}{dx} = 15x^2 + 2\).
NOTE: If \( y = c \), where \( c \) is a constant, then \( \frac{dy}{dx} = 0 \).
7.2 The Chain Rule
- Chain rule is used to differentiate composite functions.
- If \( y = f(u) \) and \( u = g(x) \), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Solving Using Chain Rule
- Differentiate the outer function with respect to the inner function.
- Differentiate the inner function with respect to \( x \).
- Multiply the results from steps 1 and 2.
Example:
For \( y = (3x^2 + 2)^4 \)
- Let \( u = 3x^2 + 2 \), so \( y = u^4 \).
- \(\frac{dy}{du} = 4u^3\)
- \(\frac{du}{dx} = 6x\)
- \(\frac{dy}{dx} = 4(3x^2 + 2)^3 \cdot 6x = 24x(3x^2 + 2)^3 \)
7.3 Tangents and Normals
Tangent
- The equation of the tangent to the curve \( y = f(x) \) at the point:
- Point \( (a, f(a)) \) :-
\[ y – f(a) = f'(a)(x – a) \] - If the value of \( \frac{dy}{dx} \) at the point \( A (x_1, y_1) \) is \( m \), then the equation of the tangent at \( A \) is given by:
\[ y – y_1 = m(x – x_1) \]
Normal
- The normal to the curve at a given point is perpendicular to the tangent.
- The gradient of the normal is \( -\frac{1}{f'(a)} \) or \( -\frac{1}{m} \).
- The equation of the normal:
\[ y – y_1 = -\frac{1}{m} (x – x_1) \]
Common Mistake: We confuse between normal and tangents.
Example:
For \( y = x^3 \) at \( x = 1 \):
- \( f'(x) = 3x^2 \), so \( f'(1) = 3 \).
- Equation of the tangent:
\[ y – 1 = 3(x – 1) \]
or
\[ y = 3x – 2 \]. - Gradient of the normal: \( -\frac{1}{3} \).
- Equation of the normal:
\[ y – 1 = -\frac{1}{3} (x – 1) \]
or
\[ y = -\frac{1}{3}x + \frac{4}{3} \].
7.4 Second Derivative
- The second derivative \( \frac{d^2y}{dx^2} \) is the derivative of the first derivative \( \frac{dy}{dx} \).
Example: Find the second derivative of \( y = (3x^2 + 4)^2 \)
- First, find the first derivative: \[ \frac{d}{dx} \left[ (3x^2 + 4)^2 \right] = 2(3x^2 + 4) \cdot \frac{d}{dx} (3x^2 + 4) \] \[ = 2(3x^2 + 4) \cdot (6x) \] \[ = 12x(3x^2 + 4) \]
- Now, find the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left[ 12x(3x^2 + 4) \right] \] \[ = 12(3x^2 + 4) + 12x(6x) \] \[ = 36x + 48 + 72x^2 \]
