What is wrong second parametric derivative?

d^2ydx^2 is not d^2y/dt^2d^2x/dt^2; you differentiate dydx with respect to t and then divide by dxdt.

For x^2 + y^2 = 25, find dydx. [2 marks]

SQA AH style: quotient rule-style practice: Differentiate y = sin xx^2 with respect to x.

Quotient rule with u = sin x, v = x^2, so u' = cos x, v' = 2x (1 mark). dydx = u'v - uv'v^2 = x^2cos x - 2xsin xx^4 (1 mark). Cancel one factor of x: dydx = xcos x - 2sin xx^3 (1 mark). Markers reward correct u, v derivatives, the quotient-rule structure, and a simplified answer.

SQA AH style: implicit-style practice: A curve has equation x^2 + xy + y^2 = 7. Find dydx and the gradient at (1, 2).

Differentiate term by term, treating y as a function of x: 2x + (y + xdydx) + 2ydydx = 0, using the product rule on xy (2 marks). Collect: dydx(x + 2y) = -(2x + y), so dydx = -2x + yx + 2y (1 mark). At (1, 2): dydx = -2 + 21 + 4 = -45 (1 mark). Markers reward the implicit differentiation including the product rule, the rearrangement, and the evaluated gradient.

ScotlandMathsSyllabus dot point

How do you differentiate products, quotients, composite, inverse, implicit and parametric functions?

Differentiate using the chain, product and quotient rules; differentiate exponential, logarithmic, inverse trigonometric, implicit and parametrically defined functions; and use logarithmic differentiation and higher derivatives.

A focused answer to the SQA Advanced Higher Mathematics differentiation techniques content, covering the chain, product and quotient rules, differentiation of exponential, logarithmic and inverse trigonometric functions, implicit and parametric differentiation, logarithmic differentiation, and the second derivative.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The three structural rules are the chain rule $\dfrac{dy}{dx} = \dfrac{dy}{du}\dfrac{du}{dx}$ , the product rule $(uv)' = u'v + uv'$ , and the quotient rule $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ . The standard derivatives you must know include $\dfrac{d}{dx}(e^x) = e^x$ , $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ , and $\dfrac{d}{dx}(\tan^{-1} x) = \dfrac{1}{1 + x^2}$ . For a curve defined implicitly, differentiate every term in $x$ , writing $\dfrac{dy}{dx}$ each time $y$ is differentiated, then solve for $\dfrac{dy}{dx}$ . For a parametric curve, $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .

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What this dot point is asking
The three rules
Standard derivatives
Logarithmic differentiation
Implicit and parametric differentiation
Try this

What this dot point is asking

The SQA wants you to differentiate anything the course can throw at you: products, quotients, composite functions, the exponential and logarithm, the inverse trigonometric functions, and curves given implicitly or parametrically. This is the engine room of the calculus, so fluency here is what frees up time in the exam.

The three rules

Every harder derivative is built from the chain, product and quotient rules. The skill is spotting which structure a function has before reaching for a rule.

Differentiate

y = x^2 e^{3x}

This function is a product of two separate functions, so the product rule $(uv)' = u'v + uv'$ is the right tool. The first step is always to identify which factor is $u$ and which is $v$ .

Step 1: Identify the structure and choose $u$ and $v$

The function is a product of $u = x^2$ and $v = e^{3x}$ . Labelling them before differentiating prevents confusion when substituting into the rule.

Step 2: Differentiate each factor

$u = x^2$ gives $u' = 2x$ by the power rule. For $v = e^{3x}$ , the chain rule is needed: the outer function is the exponential and the inner function is $3x$ , so $v' = 3e^{3x}$ .

Step 3: Apply the product rule and simplify

Substitute into $(uv)' = u'v + uv'$ , then factorise to write the answer neatly.

\dfrac{dy}{dx} = 2x\,e^{3x} + x^2\,(3e^{3x}) = e^{3x}(2x + 3x^2)

Final answer: $\dfrac{dy}{dx} = e^{3x}(2x + 3x^2)$ .

Standard derivatives

Beyond polynomials and the basic trigonometric functions from Higher, Advanced Higher adds the exponential and logarithm and, crucially, the inverse trigonometric functions.

Logarithmic differentiation

When a function is a messy product, quotient or power, taking logarithms first turns it into a sum that is easy to differentiate. This is the standard route for expressions like $y = x^x$ or a product of several factors.

Differentiate

y = x^{x}

for

x > 0

The function $x^x$ has a variable base and a variable exponent, so neither the power rule nor $\dfrac{d}{dx}(a^x) = a^x \ln a$ applies directly. Taking the natural logarithm of both sides converts the exponent into a coefficient, making the right-hand side something we can differentiate.

Step 1: Take the natural logarithm of both sides

Using $\ln(x^x) = x\ln x$ , the equation becomes:

\ln y = x\ln x

Step 2: Differentiate both sides with respect to $x$

The left side is $\dfrac{d}{dx}(\ln y) = \dfrac{1}{y}\dfrac{dy}{dx}$ by the chain rule. On the right, $x\ln x$ is a product, so use the product rule: $(x)'(\ln x) + x(\ln x)' = \ln x + 1$ .

\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + 1

Step 3: Solve for $\dfrac{dy}{dx}$ by multiplying through by $y$

Substitute $y = x^x$ to write the answer in terms of $x$ alone.

\dfrac{dy}{dx} = y(\ln x + 1) = x^{x}(\ln x + 1)

Final answer: $\dfrac{dy}{dx} = x^{x}(\ln x + 1)$ .

Implicit and parametric differentiation

When a curve is not given as $y = f(x)$ , you cannot differentiate directly. For an implicit relation, differentiate every term with respect to $x$ and attach $\dfrac{dy}{dx}$ whenever you differentiate a $y$ , then make $\dfrac{dy}{dx}$ the subject. For a parametric curve with $x = x(t)$ and $y = y(t)$ , divide the rates: $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .

Find

\dfrac{dy}{dx}

for the parametric curve

x = t^2

y = t^3 - t

When a curve is given in parametric form, neither $x$ nor $y$ is a direct function of the other. The parametric rule $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ lets us find the gradient without eliminating the parameter, by dividing the two rates.

Step 1: Differentiate each coordinate with respect to $t$

Differentiate $x$ and $y$ separately with respect to the parameter $t$ , treating each as a straightforward function.

\dfrac{dx}{dt} = 2t, \qquad \dfrac{dy}{dt} = 3t^2 - 1

Step 2: Divide the rates to get $\dfrac{dy}{dx}$

The gradient at any point is found by dividing $dy/dt$ by $dx/dt$ . The result is still in terms of $t$ , which is fine: each value of $t$ corresponds to a point on the curve.

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2 - 1}{2t}

Step 3: Evaluate at a particular value of $t$

Substitute $t = 1$ to find the gradient at that specific point on the curve.

\dfrac{dy}{dx}\bigg|_{t=1} = \dfrac{3(1)^2 - 1}{2(1)} = \dfrac{2}{2} = 1

Final answer: $\dfrac{dy}{dx} = \dfrac{3t^2 - 1}{2t}$ ; the gradient at $t = 1$ is $1$ .

For the second derivative of a parametric curve, differentiate $\dfrac{dy}{dx}$ with respect to $t$ and divide again by $\dfrac{dx}{dt}$ : $\dfrac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ . A common slip is to think you can simply differentiate $\dfrac{dy}{dx}$ with respect to $x$ directly.

Try this

Q1. Differentiate $y = \tan^{-1}(2x)$ . [2 marks]

Cue. Chain rule: $\dfrac{dy}{dx} = \dfrac{1}{1 + (2x)^2}\cdot 2 = \dfrac{2}{1 + 4x^2}$ .

Q2. For $x^2 + y^2 = 25$ , find $\dfrac{dy}{dx}$ . [2 marks]

Cue. $2x + 2y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: quotient rule3 marksDifferentiate

y = \dfrac{\sin x}{x^2}

with respect to

x

Show worked answer →

Quotient rule with $u = \sin x$ , $v = x^2$ , so $u' = \cos x$ , $v' = 2x$ (1 mark).

$\dfrac{dy}{dx} = \dfrac{u'v - uv'}{v^2} = \dfrac{x^2\cos x - 2x\sin x}{x^4}$ (1 mark).

Cancel one factor of $x$ : $\dfrac{dy}{dx} = \dfrac{x\cos x - 2\sin x}{x^3}$ (1 mark). Markers reward correct $u, v$ derivatives, the quotient-rule structure, and a simplified answer.

AH style: implicit4 marksA curve has equation

x^2 + xy + y^2 = 7

. Find

\dfrac{dy}{dx}

and the gradient at

(1, 2)

Show worked answer →

Differentiate term by term, treating $y$ as a function of $x$ : $2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$ , using the product rule on $xy$ (2 marks).

Collect: $\dfrac{dy}{dx}(x + 2y) = -(2x + y)$ , so $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$ (1 mark).

At $(1, 2)$ : $\dfrac{dy}{dx} = -\dfrac{2 + 2}{1 + 4} = -\dfrac{4}{5}$ (1 mark). Markers reward the implicit differentiation including the product rule, the rearrangement, and the evaluated gradient.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)