1.07s - Implicit and parametric differentiation | OCR A-Level Mathematics A

Can you find a tangent without first writing the curve as $y=f(x)$ ? Use 1.07s to differentiate relations between coordinates and equations in a shared parameter. Find first derivatives, tangents, normals and special tangent points while keeping the correct branch and point.

Differentiating a relation with a changing y

What is the tangent to $x^2+xy+y^2=7$ at $(1,2)$ ? Rearranging as a quadratic in $y$ would introduce a square root and a branch choice. Differentiating the relation directly is much cleaner.

You already know

The chain rule differentiates a changing input: $(y^2)'=2y\,y'$ , when $y$ depends on $x$ . The product rule gives $(xy)'=y+xy'$ . A tangent uses a numerical gradient at a point on the curve.

A letter is not constant just because it differs from $x$ . Along a differentiable branch, $y$ changes as $x$ changes.

Implicit differentiation

Implicit differentiation finds a derivative from a relation involving $x$ and $y$ , treating $y$ as a function of $x$ locally, without first isolating $y$ . For example, differentiating $x^2+y^2=25$ gives $2x+2y\,y'=0$ .

The chain and product rules used here are not in the formula sheet: learn them. Collect every term containing $y'$ before dividing.

Worked example 1: a mixed quadratic relation

Question. Differentiate $x^2+xy+y^2=7$ implicitly.

Plan. Differentiate every term with respect to $x$ , then solve the resulting equation for $y'$ .

\begin{aligned} 2x+(y+xy')+2yy'&=0 &&\text{product and chain rules}\\ (x+2y)y'&=-(2x+y) &&\text{collect }y'\\ y'&=-\frac{2x+y}{x+2y} &&\text{divide if }x+2y\ne0 \end{aligned}

Why this step? Bracketing the two product terms prevents the contribution $y$ from disappearing.

Answer. $\mathrm dy/\mathrm dx=-(2x+y)/(x+2y)$ , on branches where $x+2y\ne0$ .

Check. At $(1,2)$ , the relation gives $1+2+4=7$ and the derivative gives $-4/5$ .

Change only the coefficient of $y^2$ , from one to two.

Worked example 2: changing one quadratic coefficient

Question. Differentiate $x^2+xy+2y^2=7$ implicitly.

Plan. Keep the product calculation and update the chain-rule term.

\begin{aligned} 2x+y+xy'+4yy'&=0 &&\text{differentiate}\\ (x+4y)y'&=-(2x+y) &&\text{collect}\\ y'&=-\frac{2x+y}{x+4y} &&\text{divide if permitted} \end{aligned}

Why this step? Doubling $y^2$ doubles its derivative contribution but does not alter the $xy$ terms.

Answer. $y'=-(2x+y)/(x+4y)$ , where $x+4y\ne0$ .

Check. $(1,3/2)$ gives $1+3/2+9/2=7$ and slope $-1/2$ .

The method also gives $(\mathrm e^y)'=\mathrm e^y y'$ and $(\sin y)'=\cos y\,y'$ , with radian input. For $y+\ln y=x^2$ , $y>0$ , it gives $(1+1/y)y'=2x$ , hence $y'=2xy/(y+1)$ ; the point $(1,1)$ has slope one. In $x^2+2xy+3y^2=4$ , collecting the terms gives $y'=-(x+y)/(x+3y)$ .

Watch out

Wrong: differentiate $y^2$ as $2y$ or $\ln y$ as $1/x$ . Right: $2y\,y'$ and $y'/y$ . Both use the actual changing input.

Change just the coefficient of $y^2$ again.

Using implicit gradients and checking special tangents

A derivative expression can contain both coordinates. A line's gradient must be evaluated at the specified point, rather than left as a changing expression.

Worked example 3: tangent and normal at the opening point

Question. Find the tangent and normal to $x^2+xy+y^2=7$ at $(1,2)$ .

Plan. Check the point, evaluate the derivative, then use the point–gradient form.

\begin{aligned} 1+2+4&=7 &&\text{point is on the curve}\\ m&=-\frac{2(1)+2}{1+2(2)}=-\frac45 &&\text{tangent gradient}\\ m_{\mathrm n}&=-\frac1m=\frac54 &&\text{perpendicular gradient} \end{aligned}

Why this step? Both lines pass through the same point; only their gradients differ.

Answer. Tangent $y-2=-\tfrac45(x-1)$ ; normal $y-2=\tfrac54(x-1)$ .

Check. Both contain $(1,2)$ , and their gradient product is minus one.

Change only the chosen point on the same curve, interchanging its coordinates.

Worked example 4: a second point on the same curve

Question. Find the tangent and normal at $(2,1)$ .

Plan. Evaluate the same derivative at the new coordinate pair.

\begin{aligned} 4+2+1&=7 &&\text{check membership}\\ m&=-\frac{4+1}{2+2}=-\frac54 &&\text{evaluate}\\ m_{\mathrm n}&=\frac45 &&\text{negative reciprocal} \end{aligned}

Why this step? The curve is unchanged, but its local gradient depends on both coordinates.

Answer. Tangent $y-1=-\tfrac54(x-2)$ ; normal $y-1=\tfrac45(x-2)$ .

Check. Interchanging coordinates reflects the curve in $y=x$ , so the tangent gradients become reciprocals.

Special directions need no infinite “gradient”. On $x^2+y^2=25$ , $y'=-x/y$ : at $(0,5)$ , the tangent is $y=5$ and normal $x=0$ . At $(5,0)$ , instead use $\mathrm dx/\mathrm dy=-y/x=0$ : the tangent is $x=5$ , normal $y=0$ .

For horizontal tangents, solve the zero-numerator condition together with the curve equation and retain a nonzero denominator. The hyperbola $xy=2$ has $y'=-y/x$ , but $y=0$ contradicts its equation, so it has no horizontal tangent.

Watch out

A zero denominator alone does not establish a vertical tangent. If numerator and denominator both vanish, inspect the curve: $y^2=x^2$ has two crossing lines at the origin, not one unique tangent.

Now use the same curve at the reflected point.

Finding a gradient from two parameter rates

If $x$ and $y$ depend on a parameter $t$ , neither coordinate needs to be eliminated first. Their two rates determine the direction of the curve.

Parametric first derivative

Where the coordinate functions are differentiable and $\mathrm dx/\mathrm dt\ne0$ ,

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

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

Not in the formula sheet: learn it.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics

Statistics

Mechanics

1.07s - 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

Continue this lesson.

Learn faster with Nova