Calculus: Differentiation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your NECO exam.

Differentiation is the process of finding the derivative of a function. The derivative measures the rate of change of a function with respect to its variable.

First Principles: $$f’(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

This gives the gradient of the tangent to the curve at any point.

Standard Derivatives:

Function $f(x)$	Derivative $f’(x)$
$x^n$	$nx^{n-1}$
$k$ (constant)	$0$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\frac{1}{x}$
$e^{kx}$	$ke^{kx}$

Rules:

• Constant multiple: $\frac{d}{dx}[k \cdot f(x)] = k \cdot f’(x)$
• Sum/difference: $\frac{d}{dx}[f(x) \pm g(x)] = f’(x) \pm g’(x)$
• Product rule: $\frac{d}{dx}[f(x) \cdot g(x)] = f’(x)g(x) + f(x)g’(x)$
• Quotient rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f’(x)g(x) - f(x)g’(x)}{[g(x)]^2}$
• Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ (for composite functions)

⚡ NECO Tip: For a function like $y = (3x^2 + 5)^4$, use the chain rule: $\dfrac{dy}{dx} = 4(3x^2+5)^3 \cdot 6x = 24x(3x^2+5)^3$.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for NECO Mathematics students with a few days to months.

Derivatives of Trigonometric Functions:

$$\frac{d}{dx}(\sin x) = \cos x, \quad \frac{d}{dx}(\cos x) = -\sin x$$ $$\frac{d}{dx}(\tan x) = \sec^2 x, \quad \frac{d}{dx}(\cot x) = -\csc^2 x$$ $$\frac{d}{dx}(\sec x) = \sec x \tan x, \quad \frac{d}{dx}(\csc x) = -\csc x \cot x$$

Second Derivative: $$f”(x) = \frac{d}{dx}[f’(x)]$$ If $f’(x) = 0$ at a point and $f”(x) > 0$: local minimum. If $f’(x) = 0$ at a point and $f”(x) < 0$: local maximum.

Product Rule Examples:

Example: Differentiate $y = x^2 \sin x$ $$\frac{dy}{dx} = 2x \sin x + x^2 \cos x$$

Quotient Rule Example:

Example: Differentiate $y = \dfrac{x}{x+1}$ $$f = x, f’ = 1; g = x+1, g’ = 1$$ $$\frac{dy}{dx} = \frac{1(x+1) - x(1)}{(x+1)^2} = \frac{x+1-x}{(x+1)^2} = \frac{1}{(x+1)^2}$$

Chain Rule Examples:

Example: $y = \ln(\cos x)$ $$\frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$$

Example: $y = e^{\sin x^2}$ $$\frac{dy}{dx} = e^{\sin x^2} \cdot \cos x^2 \cdot 2x = 2x \cos x^2 \cdot e^{\sin x^2}$$

Applications of Differentiation:

1. Gradient of a curve: $f’(a)$ = gradient at $x = a$
2. Equation of tangent: $y - y_1 = m(x - x_1)$ where $m = f’(x_1)$
3. Rate of change: If $V = \frac{4}{3}\pi r^3$, then $\frac{dV}{dr} = 4\pi r^2$
4. Maximum and minimum problems: Find where $f’(x) = 0$, then check $f”(x)$

⚡ NECO Common Mistakes:

• Forgetting the chain rule when differentiating composite functions
• Using the quotient rule when product rule would be simpler (e.g., $x/(x+1)$ can be written as $x(x+1)^{-1}$)
• Getting the sign wrong in the quotient rule formula
• Mixing up local maximum/minimum with absolute maximum/minimum

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for NECO and JAMB Mathematics preparation.

Differentiation of Logarithmic Functions:

$$\frac{d}{dx}(\ln x) = \frac{1}{x}, \quad \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$$

Implicit Differentiation:

When $y$ is not isolated, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (use chain rule: $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$).

Example: Differentiate $x^2 + y^2 = 25$ $$2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$$

Differentiating Inverse Functions:

If $y = f^{-1}(x)$, then $\frac{dy}{dx} = \frac{1}{f’(y)}$ where $y = f^{-1}(x)$.

Example: Find $\frac{d}{dx}(\sin^{-1} x)$ Let $y = \sin^{-1} x \Rightarrow \sin y = x$ Differentiating: $\cos y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - \sin^2 y}} = \frac{1}{\sqrt{1 - x^2}}$

So: $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$

Related Rates:

If two variables $x$ and $y$ are related by an equation and both change with time $t$: $$\frac{dx}{dt} \text{ and } \frac{dy}{dt} \text{ are related by differentiating the equation with respect to } t$$

Example: A ladder 10 m long leans against a wall. The bottom slides away at 2 m/s. How fast is the top sliding down when the bottom is 6 m from the wall? $x^2 + y^2 = 100$. Differentiate w.r.t $t$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ When $x=6$: $y = \sqrt{100-36} = 8$. Given $\frac{dx}{dt} = 2$: $2(6)(2) + 2(8)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{24}{16} = -1.5$ m/s

Turning Points and Curve Sketching:

• Stationary point: where $f’(x) = 0$
• Point of inflection: where $f”(x) = 0$ and $f”(x)$ changes sign

For $y = ax^3 + bx^2 + cx + d$:

• $y’ = 3ax^2 + 2bx + c = 0$ gives stationary points
• $y” = 6ax + 2b$. If $y” > 0$: minimum; if $y” < 0$: maximum.

Maclaurin Series: $$f(x) = f(0) + f’(0)x + \frac{f”(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

NECO/JAMB Patterns:

• NECO frequently asks: differentiate using product, quotient, and chain rules; find equations of tangents and normals; solve maxima/minima problems; differentiate implicit functions; find second derivatives

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