Mc16, Mg15 - Implicit and parametric differentiation | OCR A-Level Mathematics B (MEI)

A relation can reveal a gradient without being rearranged into $y=f(x)$ . Mc16 develops implicit differentiation; Mg15 finds gradients from two parameter rates. Use first derivatives to match points, construct tangents and normals, and distinguish a genuine tangent from a failed derivative ratio.

Differentiating a relation with a changing y

What is the tangent to $(x+y)^2=2x$ at $(2,0)$ ? You could isolate $y=-x\pm\sqrt{2x}$ , but that immediately raises a branch question. Differentiate the relation itself, keeping the square intact.

You already know

The chain rule differentiates an outer function at its changing input. The product rule gives $(xy)'=y+xy'$ , and a tangent at a point uses the derivative evaluated there.

Along a differentiable branch, $y$ changes with $x$ , so the inner expression $x+y$ has derivative $1+y'$ .

Implicit differentiation

Implicit differentiation finds the derivative of a relation between coordinates without first isolating one coordinate. Treat $y$ as a local function of $x$ : $(y^2)'=2yy'$ , $(\mathrm e^y)'=\mathrm e^y y'$ , and $(\sin y)'=\cos y\,y'$ for radian input.

The chain and product rules are not in the formula sheet: learn them. Keep the result as an equation until all $y'$ terms have been collected.

Worked example 1: a square of a coordinate sum

Question. Show that $(x+y)^2=2x$ has derivative $y'=1/(x+y)-1$ , where $x+y\ne0$ .

Plan. Differentiate the outer square, include the full inner derivative, then isolate $y'$ .

\begin{aligned} 2(x+y)(1+y')&=2 &&\text{chain rule}\\ 1+y'&=\frac1{x+y} &&\text{divide if nonzero}\\ y'&=\frac1{x+y}-1 &&\text{rearrange} \end{aligned}

Why this step? The derivative of $x+y$ is not just one: $y$ is changing too.

Answer. The stated derivative follows for $x+y\ne0$ .

Check. At $(2,0)$ , the relation gives $4=4$ and the slope is $-1/2$ .

Change only the coefficient on the right, from two to four.

Worked example 2: the same outer structure

Question. Differentiate $(x+y)^2=4x$ .

Plan. Retain the same chain-rule left side and update the right-hand derivative.

\begin{aligned} 2(x+y)(1+y')&=4 &&\text{differentiate}\\ 1+y'&=\frac2{x+y} &&\text{divide}\\ y'&=\frac2{x+y}-1 &&\text{isolate derivative} \end{aligned}

Why this step? Only the right-hand rate changes; the outer square still multiplies the whole inner rate.

Answer. $y'=2/(x+y)-1$ , where $x+y\ne0$ .

Check. The point $(1,1)$ satisfies the relation and gives slope zero.

Expanded relations need the product rule too. For $x^2+xy+y^2=7$ ,

2x+(y+xy')+2yy'=0 \quad\Longrightarrow\quad y'=-\frac{2x+y}{x+2y}

where the denominator is nonzero. Likewise $x^2+2xy+3y^2=4$ gives $y'=-(x+y)/(x+3y)$ . A logarithmic relation $y+\ln y=x^2$ , $y>0$ , gives $(1+1/y)y'=2x$ , hence $y'=2xy/(y+1)$ ; its slope at $(1,1)$ is one.

Watch out

Wrong: $(x+y)^2$ differentiates to $2(x+y)$ . Right: $2(x+y)(1+y')$ . The missing inner factor is the changing coordinate's contribution.

Change the right-hand coefficient again.

Using implicit gradients and checking special tangents

Two points with the same $x$ -coordinate can belong to different branches. A derivative containing $y$ is useful precisely because it keeps that branch information.

Worked example 3: the upper branch at the opening point

Question. Determine the tangent and normal to $(x+y)^2=2x$ at $(2,0)$ .

Plan. Check the point, evaluate the implicit derivative, then use a negative reciprocal for the normal.

\begin{aligned} (2+0)^2&=4=2(2) &&\text{point check}\\ m&=\frac1{2+0}-1=-\frac12 &&\text{tangent gradient}\\ m_{\mathrm n}&=2 &&\text{perpendicular gradient} \end{aligned}

Why this step? The chosen $y$ -coordinate selects a branch; $x=2$ alone does not.

Answer. Tangent $y=-\tfrac12(x-2)$ ; normal $y=2(x-2)$ .

Check. Both pass through $(2,0)$ , and their gradient product is minus one.

Change only the selected branch point at $x=2$ .

Worked example 4: the lower branch at the same input

Question. Determine the tangent and normal at $(2,-4)$ on the same relation.

Plan. Use the new value of $x+y$ before finding the line equations.

\begin{aligned} (2-4)^2&=4=2(2) &&\text{point check}\\ m&=\frac1{-2}-1=-\frac32 &&\text{evaluate}\\ m_{\mathrm n}&=\frac23 &&\text{negative reciprocal} \end{aligned}

Why this step? A different branch changes the local gradient even though the horizontal coordinate is unchanged.

Answer. Tangent $y+4=-\tfrac32(x-2)$ ; normal $y+4=\tfrac23(x-2)$ .

Check. The explicit lower branch $-x-\sqrt{2x}$ also has derivative $-1-1/\sqrt{2x}$ , giving $-3/2$ at two.

The older mixed relation $x^2+xy+y^2=7$ illustrates the same point: at $(1,2)$ , tangent gradient $-4/5$ and normal gradient $5/4$ ; at $(2,1)$ , the gradients are $-5/4$ and $4/5$ . The normal lines are $y-2=\tfrac54(x-1)$ and $y-1=\tfrac45(x-2)$ .

Special directions need their own treatment. For $x^2+y^2=25$ , the point $(0,5)$ has tangent $y=5$ , normal $x=0$ . At $(5,0)$ , differentiating in the reverse direction gives $\mathrm dx/\mathrm dy=-y/x=0$ , establishing tangent $x=5$ and normal $y=0$ . On $xy=2$ , a horizontal tangent would require $y=0$ , contradicting the curve, so there is none.

Exam tip

Bracket the two product-rule terms when differentiating $xy$ . For a given point, show its substituted gradient before giving the line. A zero denominator alone does not establish a vertical tangent: a $0/0$ expression needs the curve inspected.

Use the same relation at a point where the tangent becomes horizontal.

Finding a gradient from two parameter rates

A parameter supplies both coordinates. To obtain change in $y$ per unit change in $x$ , divide their changes per parameter unit. Eliminating the parameter is optional.

Parametric first derivative

For differentiable coordinate functions $x(t),y(t)$ , where $\mathrm dx/\mathrm dt\ne0$ ,

\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}

This follows by rearranging $\mathrm dy/\mathrm dt=(\mathrm dy/\mathrm dx)(\mathrm dx/\mathrm dt)$ .

Not in the formula sheet: learn it.

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Mc16, Mg15 - Implicit and parametric differentiation

Differentiating a relation with a changing y

Using implicit gradients and checking special tangents

Finding a gradient from two parameter rates

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