Maths4U
« Back to Subchapter

Differentiation – Implicit differentiation

📘 Notes

Differentiation — Implicit Differentiation

Implicit differentiation is used when \(y\) is not written explicitly as a function of \(x\).

1. What is implicit differentiation?

Sometimes an equation contains both \(x\) and \(y\), but \(y\) has not been made the subject.

Example:

\[ x^2+y^2=25 \]

In this case, we differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\).

2. Key idea

When differentiating a term involving \(y\), we must multiply by \(\dfrac{dy}{dx}\).

\[ \frac{d}{dx}(y)=\frac{dy}{dx} \] \[ \frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx} \]

This is because \(y\) depends on \(x\), so the chain rule is needed.

3. Basic method

  1. Differentiate both sides with respect to \(x\).
  2. Remember to attach \(\dfrac{dy}{dx}\) to terms involving \(y\).
  3. Collect all \(\dfrac{dy}{dx}\) terms on one side.
  4. Factor out \(\dfrac{dy}{dx}\).
  5. Solve for \(\dfrac{dy}{dx}\).

4. Worked example 1

Find \(\dfrac{dy}{dx}\) if

\[ x^2+y^2=25 \]

Differentiate both sides:

\[ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(25) \]

\[ 2x+2y\frac{dy}{dx}=0 \]

Rearrange:

\[ 2y\frac{dy}{dx}=-2x \]

\[ \frac{dy}{dx}=-\frac{x}{y} \]

5. Worked example 2

Find \(\dfrac{dy}{dx}\) if

\[ x^3+xy+y^2=7 \]

Differentiate term by term:

\[ 3x^2+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=0 \]

Use the product rule on \(xy\):

\[ 3x^2+\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0 \]

Collect \(\dfrac{dy}{dx}\) terms:

\[ x\frac{dy}{dx}+2y\frac{dy}{dx}=-(3x^2+y) \]

Factorise:

\[ \frac{dy}{dx}=\frac{-(3x^2+y)}{x+2y} \]

6. Worked example 3

Find \(\dfrac{dy}{dx}\) if

\[ y^3+2x^2y=4 \]

Differentiate both sides:

\[ 3y^2\frac{dy}{dx}+2\frac{d}{dx}(x^2y)=0 \]

Use the product rule on \(x^2y\):

\[ 3y^2\frac{dy}{dx}+2\left(x^2\frac{dy}{dx}+2xy\right)=0 \]

\[ 3y^2\frac{dy}{dx}+2x^2\frac{dy}{dx}+4xy=0 \]

Factorise:

\[ \left(3y^2+2x^2\right)\frac{dy}{dx}=-4xy \]

\[ \frac{dy}{dx}=\frac{-4xy}{3y^2+2x^2} \]

7. Finding the gradient at a point

After finding \(\dfrac{dy}{dx}\), substitute the coordinates of the point.

Example: For

\[ x^2+y^2=25, \]

find the gradient at \((3,4)\).

Since \[ \frac{dy}{dx}=-\frac{x}{y}, \]

\[ \frac{dy}{dx}=-\frac{3}{4} \]

8. Second derivatives

Sometimes you may need to differentiate \(\dfrac{dy}{dx}\) again to find \(\dfrac{d^2y}{dx^2}\).

This often requires quotient rule, product rule, and implicit differentiation again.

9. Common mistakes

  • Forgetting \(\dfrac{dy}{dx}\) when differentiating \(y\)-terms.
  • Forgetting the product rule in terms like \(xy\) or \(x^2y\).
  • Stopping before making \(\dfrac{dy}{dx}\) the subject.
  • Substituting a point too early before finding the general derivative.

10. Exam tips

  • Treat \(y\) as a function of \(x\).
  • Every time you differentiate a \(y\)-term, include \(\dfrac{dy}{dx}\).
  • Look carefully for product rule if \(x\) and \(y\) are multiplied together.
  • Collect all \(\dfrac{dy}{dx}\) terms on one side before solving.
  • Write the final answer clearly in terms of \(x\) and \(y\).