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CIE A-Level Maths Study Notes

1.7.2 Differentiation Techniques

In this section, our focus will be on differentiating powers of x with rational exponents, as well as employing the chain rule for composite functions. These techniques are essential for understanding how functions change and evolve.

Differentiating Powers of x with Rational Exponents

When working with x raised to a rational exponent, the differentiation process follows a standard rule. For f(x)=xnf(x) = x^n, where nn is a rational number, the derivative is f(x)=nxn1 f'(x) = nx^{n-1}.

Example

Differentiate y=x3/2y = x^{3/2}.

Solution:

y=x3/2y = x^{3/2}dydx=32x321\frac{dy}{dx} = \frac{3}{2} x^{\frac{3}{2} - 1} =32x1/2= \frac{3}{2} x^{1/2}

Differentiation of Sums, Differences, and Constant Multiples of Functions

These rules enable simplification before applying differentiation:

1. Sum: ddx(u(x)+v(x))=u(x)+v(x)\frac{d}{dx}(u(x) + v(x)) = u'(x) + v'(x)

2. Difference: ddx(u(x)v(x))=u(x)v(x)\frac{d}{dx}(u(x) - v(x)) = u'(x) - v'(x)

3. Constant Multiple: ddx(cu(x))=cu(x)\frac{d}{dx}(c \cdot u(x)) = c \cdot u'(x)

Example

Differentiate y=3x2+4xy = 3x^2 + 4x.

Solution:

y=3x2+4xy = 3x^2 + 4x dydx=32x+41\frac{dy}{dx} = 3 \cdot 2x + 4 \cdot 1=6x+4= 6x + 4

Chain Rule

This rule is used for differentiating composite functions and is formulated as:

dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

Or,

f(g(x)))=f(g(x))×g(x)f(g(x)))' = f'(g(x)) \times g'(x)

Example

Differentiate y=(x+x3)5y = (x + x^3)^5.

Solution:

Let u=x+x3u = x + x^3, then find dudx\frac{du}{dx}

u=x+x3u = x + x^3dudx=1+3x2\frac{du}{dx} = 1 + 3x^2

Now ( y = u^5 )

dydu=5u4\frac{dy}{du} = 5u^4

Multiply them together:

dydx=dydu×dudx=(1+3x2)×5(x+x3)4\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = (1 + 3x^2) \times 5(x + x^3)^4

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