Differential Calculus Techniques

Leaving Cert Higher Level Mathematics revision notes with diagrams, key terms and self-check questions.

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Differential calculus measures the instantaneous rate of change of a function, which gives the exact slope of the tangent line at any point along its curve. For Leaving Certificate Higher Level, you build on the foundational limit definition—known as differentiation from first principles—and master the standard operational rules. These include the product, quotient, and chain rules applied across polynomials, rational powers, exponential terms, logarithms, trigonometric functions, and inverse trigonometric expressions.

Differentiation from First Principles

For a straight line, the slope never changes. A curve is different because its steepness varies from point to point. To measure the slope of a curve at a single point P(x,f(x))P(x, f(x)), we pick a neighbouring point Q(x+h,f(x+h))Q(x+h, f(x+h)) and construct a secant line between them. The slope of this secant line is the difference quotient:

m=f(x+h)f(x)hm = \frac{f(x+h) - f(x)}{h}

As you shrink hh towards zero, QQ slides along the curve towards PP. The secant line gradually pivots until it becomes the tangent line at PP. Taking the limit as h0h \to 0 gives the instantaneous rate of change, which is the derivative:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
A secant joins P and Q on a curve. Its horizontal run is h and vertical rise is f(x+h)−f(x). As Q approaches P, the secant approaches the tangent.
A secant joins P and Q on a curve. Its horizontal run is h and vertical rise is f(x+h)−f(x). As Q approaches P, the secant approaches the tangent.

The syllabus specifically requires you to differentiate linear and quadratic functions using this first principles method. On the exam, write out the limit definition clearly, expand f(x+h)f(x+h) carefully in brackets, subtract f(x)f(x), divide every remaining term by hh, and then evaluate the limit by setting h=0h = 0.

Polynomials and Rational Powers

Once the principles are established, standard rules let us differentiate power terms quickly. For any term y=axny = a x^n, multiply by the existing index and reduce the power by one:

ddx(axn)=anxn1\frac{d}{dx}(a x^n) = a n x^{n-1}

The derivative of any isolated constant is zero because a constant value has a rate of change of zero. When a function is made up of sums and differences, differentiate each term separately.

Before differentiating roots or fractions, always convert them into index form. For instance, write x\sqrt{x} as x12x^{\frac{1}{2}} and 1x3\frac{1}{x^3} as x3x^{-3}. When subtracting 1 from a negative or fractional power, take care with the arithmetic: 21=3-2 - 1 = -3, and 121=12\frac{1}{2} - 1 = -\frac{1}{2}. For example, differentiating f(x)=4x=4x12f(x) = \frac{4}{\sqrt{x}} = 4x^{-\frac{1}{2}} gives f(x)=4(12)x32=2x32f'(x) = 4\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = -\frac{2}{x^{\frac{3}{2}}}.

Product and Quotient Rules

When two functions are multiplied or divided, you cannot simply multiply or divide their individual derivatives. You must apply dedicated formulas, both of which appear on page 25 of the Formulae and Tables booklet.

The Product Rule applies to products of functions y=uvy = u v:

dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

In plain words, take the first function times the derivative of the second, plus the second function times the derivative of the first.

The Quotient Rule applies to algebraic fractions y=uvy = \frac{u}{v}:

dydx=vdudxudvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

Order matters here because of the subtraction in the numerator. It must always be the bottom times the derivative of the top, minus the top times the derivative of the bottom, all over the bottom squared. Writing out uu, vv, dudx\frac{du}{dx}, and dvdx\frac{dv}{dx} separately at the side of your page keeps your working clean and helps avoid sign slips.

Chain Rule

The chain rule handles composite functions, which are functions nested inside other functions, written as y=f(g(x))y = f(g(x)). If y=u(v)y = u(v), where v=g(x)v = g(x), the rule states:

dydx=dudv×dvdx\frac{dy}{dx} = \frac{du}{dv} \times \frac{dv}{dx}

Think of it as differentiating the outside layer while leaving the inside untouched, then multiplying by the derivative of whatever was inside. For brackets raised to a power, [g(x)]n[g(x)]^n, this gives the quick working rule:

ddx[g(x)]n=n[g(x)]n1×g(x)\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \times g'(x)

For example, to differentiate y=(3x25)4y = (3x^2 - 5)^4, treat the bracket as your variable first to get 4(3x25)34(3x^2 - 5)^3, then multiply by the internal derivative 6x6x, giving dydx=24x(3x25)3\frac{dy}{dx} = 24x(3x^2 - 5)^3.

The example passes from x to v=3x²−5 and then to y=v⁴. The two stages contribute derivative factors 6x and 4v³.
The example passes from x to v=3x²−5 and then to y=v⁴. The two stages contribute derivative factors 6x and 4v³.

Trigonometric, Exponential, and Logarithmic Functions

Trigonometric, exponential, and logarithmic functions each have a standard derivative, which you then combine with the chain, product, and quotient rules.

For trigonometric functions, your angles must be in radians:

  • ddx(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x
  • ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x
  • ddx(tanx)=sec2x\frac{d}{dx}(\tan x) = \sec^2 x
Aligned sine and cosine graphs show that rising sine corresponds to positive cosine, horizontal sine tangents to zero cosine, and falling sine to negative cosine.
Aligned sine and cosine graphs show that rising sine corresponds to positive cosine, horizontal sine tangents to zero cosine, and falling sine to negative cosine.

Applying the chain rule to an angle like kxkx introduces a factor of kk, so ddx(cos4x)=4sin4x\frac{d}{dx}(\cos 4x) = -4\sin 4x.

The natural exponential function exe^x is unique because it equals its own derivative: ddx(ex)=ex\frac{d}{dx}(e^x) = e^x. By the chain rule, whenever the power is a function f(x)f(x):

ddx(ef(x))=f(x)ef(x)\frac{d}{dx}\left(e^{f(x)}\right) = f'(x) e^{f(x)}

For the general exponential axa^x, the derivative is axlnaa^x \ln a.

The natural logarithm lnx\ln x has the derivative 1x\frac{1}{x}. When taking the log of a function, the chain rule gives:

ddx(ln(f(x)))=f(x)f(x)\frac{d}{dx}(\ln(f(x))) = \frac{f'(x)}{f(x)}

For x>1x > 1, if you face a log expression like lnx+1x1\ln\sqrt{\frac{x+1}{x-1}}, expand it using log laws into 12[ln(x+1)ln(x1)]\frac{1}{2}[\ln(x+1) - \ln(x-1)] before differentiating. It saves considerable time and prevents messy fraction work.

Inverse Trigonometric Functions and Second Derivative

For a>0a > 0, the derivatives of the inverse trigonometric functions sin1(xa)\sin^{-1}\left(\frac{x}{a}\right) and tan1(xa)\tan^{-1}\left(\frac{x}{a}\right) are given in your tables:

ddx(sin1xa)=1a2x2\frac{d}{dx}\left(\sin^{-1}\frac{x}{a}\right) = \frac{1}{\sqrt{a^2 - x^2}}ddx(tan1xa)=aa2+x2\frac{d}{dx}\left(\tan^{-1}\frac{x}{a}\right) = \frac{a}{a^2 + x^2}

Notice that sin1x\sin^{-1} x is only defined when 1x1-1 \le x \le 1, and its derivative has a square root in the denominator. The derivative of cos1(xa)\cos^{-1}\left(\frac{x}{a}\right) matches inverse sine but carries a negative sign: 1a2x2-\frac{1}{\sqrt{a^2 - x^2}}.

The second derivative, written as f(x)f''(x) or d2ydx2\frac{d^2y}{dx^2}, is simply the derivative of the first derivative. It describes how the slope changes, which controls the curve's concavity. Points where f(x)=0f''(x) = 0 (and where concavity changes sign) locate points of inflection on the graph.

A schematic increasing curve flattens and then steepens through an inflection. Its aligned second derivative changes from negative to positive.
A schematic increasing curve flattens and then steepens through an inflection. Its aligned second derivative changes from negative to positive.

Tangent to a Circle

You can find the slope of a tangent to a circle centered at the origin, x2+y2=r2x^2 + y^2 = r^2, using calculus rather than coordinate geometry. Differentiating both sides with respect to xx means treating yy as an implicit function of xx.

Differentiating x2x^2 gives 2x2x. Applying the chain rule to y2y^2 produces 2ydydx2y \frac{dy}{dx}. The constant r2r^2 becomes 00:

2x+2ydydx=0    dydx=xy2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

At a point (x1,y1)(x_1, y_1) on the circle with y10y_1 \ne 0, the slope of the tangent line is m=x1y1m = -\frac{x_1}{y_1}. You can then write the equation of the tangent line directly using yy1=m(xx1)y - y_1 = m(x - x_1). When y1=0y_1 = 0, the tangent is vertical and its equation is x=x1x = x_1.

An origin-centred circle has a tangent perpendicular to the radius at P(x₁,y₁), with slope −x₁/y₁. At the rightmost point, the tangent is vertical.
An origin-centred circle has a tangent perpendicular to the radius at P(x₁,y₁), with slope −x₁/y₁. At the rightmost point, the tangent is vertical.

Key terms

First principles
The formal limit-based method limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} used to establish the derivative of linear and quadratic functions.
Product rule
The rule for differentiating two multiplying functions: ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}.
Quotient rule
The rule for differentiating algebraic fractions: ddx(uv)=vdudxudvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}.
Chain rule
The method for differentiating composite functions by multiplying the derivative of the outer expression by the derivative of the inner expression.
Stationary point
A point on a curve where the first derivative equals zero (f(x)=0f'(x) = 0), identifying a local maximum, local minimum, or horizontal point of inflection.
Second derivative
The rate of change of the first derivative, written as f(x)f''(x) or d2ydx2\frac{d^2y}{dx^2}, used to assess curve concavity and locate points of inflection.

Check yourself

  1. What is the derivative of f(x)=6xf(x) = \frac{6}{\sqrt{x}}?

    Rewrite as f(x)=6x12f(x) = 6x^{-\frac{1}{2}}. Differentiating gives f(x)=6(12)x32=3x32f'(x) = 6\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} = -3x^{-\frac{3}{2}} or 3xx-\frac{3}{x\sqrt{x}}.

  2. What is the derivative of y=ln(5x2+1)y = \ln(5x^2 + 1)?

    Using ddx(ln(g(x)))=g(x)g(x)\frac{d}{dx}(\ln(g(x))) = \frac{g'(x)}{g(x)}, differentiate the inner expression to get 10x10x, giving dydx=10x5x2+1\frac{dy}{dx} = \frac{10x}{5x^2 + 1}.

  3. What is the derivative of y=x2cos3xy = x^2 \cos 3x?

    Use the product rule with u=x2u = x^2 and v=cos3xv = \cos 3x. With u=2xu' = 2x and v=3sin3xv' = -3\sin 3x, we get dydx=x2(3sin3x)+(cos3x)(2x)=2xcos3x3x2sin3x\frac{dy}{dx} = x^2(-3\sin 3x) + (\cos 3x)(2x) = 2x\cos 3x - 3x^2\sin 3x.

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