Implicit differentiation: derivatives, tangents and stationary points

Implicit differentiation finds derivatives when $x$ and $y$ are related by an equation that has not, or cannot easily, be rearranged into $y=f(x)$. The central idea is simple: differentiate every term with respect to $x$, remembering that $y$ is itself a function of $x$.

Before you begin

You should be confident with:

• differentiating standard functions
• the product, quotient and chain rules
• rearranging equations and substituting coordinates

Why does $\dfrac{dy}{dx}$ appear?

Suppose $y$ depends on $x$. Differentiating $y^2$ with respect to $x$ requires the chain rule:

$$\frac{d}{dx}(y^2) =\frac{d}{dy}(y^2)\frac{dy}{dx} =2y\frac{dy}{dx}.$$

The factor $\dfrac{dy}{dx}$ records the fact that a change in $x$ causes a change in $y$. More generally,

$$\frac{d}{dx}\bigl(f(y)\bigr)=f'(y)\frac{dy}{dx}.$$

Useful cases include

$$\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}, \qquad \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx},$$

and

$$\frac{d}{dx}(e^y)=e^y\frac{dy}{dx}.$$

By contrast, $\dfrac{d}{dx}(x^2)=2x$ because $x$ is already the variable with respect to which we are differentiating.

The method

For an equation involving $x$ and $y$:

1. Differentiate both sides with respect to $x$.
2. Apply the chain rule to every expression involving $y$.
3. Use the product rule if $x$ and $y$ are multiplied together.
4. Collect every term containing $\dfrac{dy}{dx}$.
5. Factor out $\dfrac{dy}{dx}$ and rearrange.

Example 1: a circle

Find $\dfrac{dy}{dx}$ for

$x^2+y^2=25.$

Differentiate each term with respect to $x$:

$$2x+2y\frac{dy}{dx}=0.$$

Now isolate $\dfrac{dy}{dx}$:

$$2y\frac{dy}{dx}=-2x,$$ $$\boxed{\frac{dy}{dx}=-\frac{x}{y}}.$$

This result makes geometric sense. At $(3,4)$ the radius from the origin has gradient $4/3$, so the tangent, which is perpendicular to the radius, has gradient $-3/4$.

Example 2: collect several derivative terms

Given

$x^2+xy+y^2=7,$

find $\dfrac{dy}{dx}$.

The middle term needs the product rule:

$$\frac{d}{dx}(xy)=x\frac{dy}{dx}+y.$$

Therefore

$$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0.$$

Collect the terms containing $\dfrac{dy}{dx}$:

$$(x+2y)\frac{dy}{dx}=-(2x+y).$$

Hence

$$\boxed{\frac{dy}{dx}=-\frac{2x+y}{x+2y}}.$$

Do not treat $xy$ as though $y$ were constant. Both factors vary with $x$.

Example 3: a chain inside a chain

Find $\dfrac{dy}{dx}$ if

$e^y+x^2y=4.$

Differentiate:

$$e^y\frac{dy}{dx} +\left(2xy+x^2\frac{dy}{dx}\right)=0.$$

Collect and factor:

$$\left(e^y+x^2\right)\frac{dy}{dx}=-2xy.$$

Therefore

$$\boxed{\frac{dy}{dx}=-\frac{2xy}{e^y+x^2}}.$$

Tangents and normals

Once $\dfrac{dy}{dx}$ is known, substitute the given point to obtain the tangent gradient. A line through $(x_1,y_1)$ with gradient $m$ has equation

$y-y_1=m(x-x_1).$

If the tangent gradient is $m\ne 0$, the normal gradient is

$m_{\text{normal}}=-\frac{1}{m}.$

Example 4: tangent and normal equations

The curve

$x^2+xy+y^2=7$

passes through $(1,2)$. From Example 2,

$$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$

At $(1,2)$,

$$m_{\text{tangent}} =-\frac{2(1)+2}{1+2(2)} =-\frac45.$$

The tangent is

$$\boxed{y-2=-\frac45(x-1)}.$$

The normal gradient is $5/4$, so the normal is

$$\boxed{y-2=\frac54(x-1)}.$$

Keep the equations in point-gradient form unless another form is requested. It reduces avoidable algebra.

Stationary points

At a stationary point, the tangent is horizontal, so

$\frac{dy}{dx}=0.$

If

$$\frac{dy}{dx}=\frac{N(x,y)}{D(x,y)},$$

then a stationary point normally satisfies $N(x,y)=0$, provided $D(x,y)\ne 0$. You must combine this condition with the original curve equation.

Example 5: locating stationary points

Find the stationary points of

$x^2+xy+y^2=7.$

We have

$$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$

For a stationary point,

$2x+y=0,$

so $y=-2x$. Substitute this into the original equation:

$$x^2+x(-2x)+(-2x)^2=7,$$ $$3x^2=7.$$

Thus

$$x=\pm\sqrt{\frac73}.$$

Since $y=-2x$, the stationary points are

$$\boxed{ \left(\sqrt{\frac73},-2\sqrt{\frac73}\right) \quad\text{and}\quad \left(-\sqrt{\frac73},2\sqrt{\frac73}\right) }.$$

At these points $x+2y=-3x\ne0$, so the derivative is defined.

Second derivatives

For an implicitly defined curve, differentiate the first derivative equation again. Do not assume $y$ becomes constant on the second differentiation.

Example 6: finding and using $\dfrac{d^2y}{dx^2}$

For the circle

$x^2+y^2=25,$

the first differentiated equation is

$2x+2y\frac{dy}{dx}=0.$

Differentiate this equation again:

$$2+2\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0.$$

The square appears because the product rule gives

$$\frac{d}{dx}\left(y\frac{dy}{dx}\right) =\frac{dy}{dx}\frac{dy}{dx}+y\frac{d^2y}{dx^2}.$$

Rearranging,

$$\boxed{ \frac{d^2y}{dx^2} =-\frac{1+\left(\frac{dy}{dx}\right)^2}{y} }.$$

At $(0,5)$, $\dfrac{dy}{dx}=0$, so

$$\frac{d^2y}{dx^2}=-\frac15<0.$$

Therefore $(0,5)$ is a local maximum of the upper branch of the circle. At $(0,-5)$, the second derivative is $1/5>0$, so the lower branch has a local minimum there.

You can also differentiate an explicit formula for $\dfrac{dy}{dx}$, but this often introduces a quotient rule. Differentiating the simpler first-derivative equation is usually cleaner.

A compact general result

If a curve is written as

$F(x,y)=0,$

then, where $F_y\ne0$,

$$\boxed{\frac{dy}{dx}=-\frac{F_x}{F_y}}.$$

Here $F_x$ means differentiate with respect to $x$ while holding $y$ constant, and $F_y$ means differentiate with respect to $y$ while holding $x$ constant. This formula explains why the numerator and denominator in an implicit derivative often resemble partial derivatives. At A level, the step-by-step method is normally clearer and earns the method marks.

Common mistakes

Missing the chain-rule factor

Incorrect:

$\frac{d}{dx}(y^3)=3y^2.$

Correct:

$\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}.$

Forgetting the product rule

Incorrect:

$\frac{d}{dx}(xy)=x\frac{dy}{dx}.$

Correct:

$\frac{d}{dx}(xy)=y+x\frac{dy}{dx}.$

Substituting the point too early

Differentiate and rearrange symbolically first. Early substitution can hide terms and makes the method harder to check.

Setting the whole fraction equal to zero without checking it

A fraction is zero when its numerator is zero and its denominator is non-zero. A zero denominator may represent a vertical tangent or a point at which the derivative formula is undefined.

Using the negative reciprocal in the wrong place

$\dfrac{dy}{dx}$ gives the tangent gradient. Take the negative reciprocal only when finding the normal gradient.

Self-check

1. Given $x^2+3y^2=12$, find $\dfrac{dy}{dx}$.
2. Given $x^2y+y^3=10$, find $\dfrac{dy}{dx}$.
3. Find the tangent to $x^2+xy+y^2=7$ at $(2,1)$.
4. For $x^2+4y^2=16$, find the stationary points and classify them.

Answers

1. $\displaystyle \frac{dy}{dx}=-\frac{x}{3y}$.
2. $\displaystyle \frac{dy}{dx}=-\frac{2xy}{x^2+3y^2}$.
3. Since $\displaystyle \frac{dy}{dx}=-\frac{2x+y}{x+2y}$, the gradient at $(2,1)$ is $-5/4$. The tangent is $\displaystyle y-1=-\frac54(x-2)$.
4. $\displaystyle \frac{dy}{dx}=-\frac{x}{4y}$, so stationary points require $x=0$. The curve gives $(0,2)$ and $(0,-2)$. Differentiating $2x+8y\dfrac{dy}{dx}=0$ again shows $\dfrac{d^2y}{dx^2}=-1/(4y)$ at a stationary point. Hence $(0,2)$ is a local maximum and $(0,-2)$ is a local minimum.

What to learn next

• Use implicit derivatives in tangents and normals.
• Compare this method with parametric differentiation.
• Apply related derivatives to connected rates of change.
• Develop classification skills in stationary points and curve sketching.