IB Mathematics: Analysis and Approaches HL — Topic 5 Calculus

Implicit Differentiation

Differentiate relations involving both \(x\) and \(y\), then use the result for tangents, normals, special tangents and higher derivatives.

AA HL · AHL 5.14

Learning goal

Differentiate equations in which \(y\) is not isolated, solve reliably for \(dy/dx\), and use the result in geometric problems.

Syllabus link

AHL 5.14 Implicit differentiation, related rates and optimisation.

Big idea

Regard \(y\) as a function of \(x\). Every differentiated \(y\)-expression gains a chain-rule factor \(dy/dx\).

One idea to remember

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

The one idea to remember

If \(y=y(x)\), then \(y\) changes when \(x\) changes. So

\[\begin{gathered}\dfrac{d}{dx}(y^2)=2y\dfrac{dy}{dx}, \\ \dfrac{d}{dx}(\sin y)=\cos y\,\dfrac{dy}{dx}, \\ \dfrac{d}{dx}(e^y)=e^y\dfrac{dy}{dx}.\end{gathered}\]

The extra factor \(\dfrac{dy}{dx}\) is simply the chain rule.

Why do we need implicit differentiation?

Many curves are most naturally given by an equation that mixes \(x\) and \(y\). For example,

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

describes a circle centred at the origin with radius 5. We could solve for \(y\), but this immediately creates two branches:

\[\begin{gathered}y=\sqrt{25-x^2} \\\text{or}\\ y=-\sqrt{25-x^2}.\end{gathered}\]

Implicit differentiation lets us differentiate the original relation directly, without choosing a branch first.

Circle x squared plus y squared equals 25, with tangent and normal at the point (3, 4)

At the point \((3,4)\), implicit differentiation will give the gradient of the tangent without ever having to write \(y\) explicitly as a function of \(x\).

The chain rule is doing the work

To see why the method works, write \(y\) as \(y(x)\). Then

\[\begin{gathered}\dfrac{d}{dx}\bigl(y^3\bigr) =\dfrac{d}{dx}\bigl([y(x)]^3\bigr) \\=3[y(x)]^2\,y'(x) \\=3y^2\dfrac{dy}{dx}.\end{gathered}\]

Expression	Derivative with respect to \(x\)
\(y^n\)	\(n y^{n-1}\dfrac{dy}{dx}\)
\(\sin y\)	\(\cos y\dfrac{dy}{dx}\)
\(\cos y\)	\(-\sin y\dfrac{dy}{dx}\)
\(e^y\)	\(e^y\dfrac{dy}{dx}\)
\(\ln y\)	\(\dfrac{1}{y}\dfrac{dy}{dx}\)
\(xy\)	\(x\dfrac{dy}{dx}+y\) (product rule)

Common trap

Do not treat \(y\) as a constant. For example,

\[\dfrac{d}{dx}(y^2)\neq 2y.\]

The correct derivative is \(2y\dfrac{dy}{dx}\), because \(y\) depends on \(x\).

A reliable four-step method

Implicit differentiation routine

Differentiate both sides with respect to \(x\). Apply every ordinary rule you already know: power, product, quotient and chain rules.
Attach \(\dfrac{dy}{dx}\) whenever differentiating a \(y\)-expression.
Collect all terms containing \(\dfrac{dy}{dx}\) on one side.
Factor out \(\dfrac{dy}{dx}\), then divide to isolate it.

Your final derivative will often contain both \(x\) and \(y\). That is completely normal.

Worked example 1: the circle \(x^2+y^2=25\)

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

Solution

Differentiate both sides:

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

Collect the derivative term:

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

Divide by \(2y\):

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

At \((3,4)\), the gradient is

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

Worked example 2: a product containing \(x\) and \(y\)

For the curve

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

find \(\dfrac{dy}{dx}\).

Solution

Differentiate term by term. The middle term requires the product rule:

\[2x+\dfrac{d}{dx}(xy)+2y\dfrac{dy}{dx}=0,\]

\[2x+\left(x\dfrac{dy}{dx}+y\right)+2y\dfrac{dy}{dx}=0.\]

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

\[(x+2y)\dfrac{dy}{dx}=-(2x+y).\]

Therefore

\[\boxed{\dfrac{dy}{dx}=-\frac{2x+y}{x+2y}}.\]

More involved algebraic relations

Worked example 3: product rule on both variables

For

\[x^3+y^3=6xy,\]

find \(\dfrac{dy}{dx}\).

Solution

Differentiate both sides:

\[3x^2+3y^2\dfrac{dy}{dx}=6\left(y+x\dfrac{dy}{dx}\right).\]

Expand and collect the derivative terms:

\[3y^2\dfrac{dy}{dx}-6x\dfrac{dy}{dx}=6y-3x^2.\]

Factor:

\[(3y^2-6x)\dfrac{dy}{dx}=6y-3x^2.\]

Hence

\[\boxed{\dfrac{dy}{dx}=\frac{2y-x^2}{y^2-2x}}.\]

A useful habit

When you differentiate, do not try to isolate \(\dfrac{dy}{dx}\) after every line. First differentiate the entire relation correctly. Then gather all \(\dfrac{dy}{dx}\)-terms together. This greatly reduces sign errors.

Implicit differentiation with composite functions

The same chain-rule principle applies when \(y\) sits inside a trigonometric, exponential or logarithmic function.

Worked example 4: \(\sin(x+y)=xy\)

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

Solution

The left side is a composite function. Differentiate \(\sin(\,\cdot\,)\), then multiply by the derivative of \(x+y\):

\[\cos(x+y)\left(1+\dfrac{dy}{dx}\right)=y+x\dfrac{dy}{dx}.\]

Expand:

\[\begin{gathered}\cos(x+y)+\cos(x+y)\dfrac{dy}{dx}\\=y+x\dfrac{dy}{dx}.\end{gathered}\]

Collect the derivative terms:

\[\begin{gathered}\bigl(\cos(x+y)-x\bigr)\dfrac{dy}{dx}\\=y-\cos(x+y).\end{gathered}\]

Therefore

\[\boxed{\dfrac{dy}{dx}=\frac{y-\cos(x+y)}{\cos(x+y)-x}}.\]

Worked example 5: an exponential containing \(xy\)

For

\[e^{xy}+y^2=4,\]

find \(\dfrac{dy}{dx}\).

Solution

The exponent \(xy\) also depends on \(x\), so the chain rule and product rule are both needed:

\[e^{xy}\dfrac{d}{dx}(xy)+2y\dfrac{dy}{dx}=0,\]

\[e^{xy}\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0.\]

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

\[\left(xe^{xy}+2y\right)\dfrac{dy}{dx}=-ye^{xy}.\]

Thus

\[\boxed{\dfrac{dy}{dx}=-\frac{ye^{xy}}{xe^{xy}+2y}}.\]

Worked example 6: a logarithmic relation

For

\[\ln(x^2+y^2)=x,\]

find \(\dfrac{dy}{dx}\).

Solution

Differentiate using the chain rule:

\[\frac{2x+2y\dfrac{dy}{dx}}{x^2+y^2}=1.\]

Multiply through by \(x^2+y^2\):

\[2x+2y\dfrac{dy}{dx}=x^2+y^2.\]

Therefore

\[\boxed{\dfrac{dy}{dx}=\frac{x^2+y^2-2x}{2y}}.\]

Tangents and normals to implicitly defined curves

Once \(\dfrac{dy}{dx}\) is known, tangent and normal questions work exactly as they did for explicit functions.

Line gradients

At a point \((x_0,y_0)\):

\[\begin{gathered}m_{\text{tangent}}=\left.\dfrac{dy}{dx}\right|_{(x_0,y_0)}, \\ m_{\text{normal}}=-\frac{1}{m_{\text{tangent}}},\end{gathered}\]

provided the tangent gradient is finite and non-zero.

Worked example 7: tangent and normal

The point \((1,2)\) lies on

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

Find the equations of the tangent and normal at this point.

Solution

From Worked example 2,

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

At \((1,2)\),

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

So the tangent is

\[y-2=-\frac45(x-1),\]

\[\boxed{4x+5y=14}.\]

The normal gradient is the negative reciprocal:

\[m_{\text{normal}}=\frac54.\]

Hence the normal is

\[y-2=\frac54(x-1),\]

\[\boxed{5x-4y+3=0}.\]

Always check the point

Before substituting a point into an implicit derivative, verify that the point actually lies on the curve. In an exam question this may be automatic, but in algebraic work it is an excellent error check.

Horizontal and vertical tangents

Suppose the derivative has the form

\[\dfrac{dy}{dx}=\frac{N(x,y)}{D(x,y)}.\]

Then:

a horizontal tangent usually occurs when \(N(x,y)=0\) and \(D(x,y)\neq0\);
a vertical tangent usually occurs when \(D(x,y)=0\) and \(N(x,y)\neq0\).

The point must also satisfy the original curve equation.

Worked example 8: horizontal tangents

Find the points where

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

has a horizontal tangent.

Solution

We have

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

For a horizontal tangent,

\[\begin{gathered}2x+y=0 \\\quad\Rightarrow\quad y=-2x.\end{gathered}\]

Substitute into the original curve:

\[x^2+x(-2x)+(-2x)^2=7,\]

\[3x^2=7.\]

Thus

\[\begin{gathered}x=\pm\frac{\sqrt{21}}{3},\\ y=-2x.\end{gathered}\]

Therefore the two points are

\[\begin{gathered}\boxed{\begin{gathered}\left(\frac{\sqrt{21}}3,-\frac{2\sqrt{21}}3\right),\\ \left(-\frac{\sqrt{21}}3,\frac{2\sqrt{21}}3\right)\end{gathered}}.\end{gathered}\]

Worked example 9: vertical tangents

Find the points on \(x^2+xy+y^2=7\) where the tangent is vertical.

Solution

A vertical tangent occurs when

\[\begin{gathered}x+2y=0 \\\quad\Rightarrow\quad x=-2y.\end{gathered}\]

Substitute into the curve:

\[(-2y)^2+(-2y)y+y^2=7,\]

\[3y^2=7.\]

Hence

\[\begin{gathered}y=\pm\frac{\sqrt{21}}3,\\ x=-2y.\end{gathered}\]

So the vertical tangents occur at

\[\begin{gathered}\boxed{\begin{gathered}\left(-\frac{2\sqrt{21}}3,\frac{\sqrt{21}}3\right),\\ \left(\frac{2\sqrt{21}}3,-\frac{\sqrt{21}}3\right)\end{gathered}}.\end{gathered}\]

Second derivatives implicitly

Implicit differentiation can be repeated. Differentiate the first-derivative relation again, remembering that \(\dfrac{dy}{dx}\) is itself a function of \(x\).

Worked example 10: finding \(\dfrac{d^2y}{dx^2}\) on a circle

For

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

find \(\dfrac{d^2y}{dx^2}\), and evaluate it at \((3,4)\).

Solution

We obtained

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

Differentiate this equation again:

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

Use the product rule on \(y\dfrac{dy}{dx}\):

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

\[y\dfrac{d^2y}{dx^2}=-1-\left(\dfrac{dy}{dx}\right)^2,\]

and therefore

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

At \((3,4)\), \(\dfrac{dy}{dx}=-\dfrac34\), so

\[\dfrac{d^2y}{dx^2}=-\frac{1+\frac{9}{16}}{4}=-\frac{25}{64}.\]

Why not solve for \(y\) first?

Sometimes it is possible to rearrange an implicit equation before differentiating, but this is often less efficient and can force us to choose between different branches of the curve.

For the circle

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

solving explicitly gives

\[y=\pm\sqrt{25-x^2}.\]

If we differentiate the upper branch,

\[y=\sqrt{25-x^2},\]

then

\[\begin{gathered}\dfrac{dy}{dx}=\frac{-2x}{2\sqrt{25-x^2}}\\=-\frac{x}{\sqrt{25-x^2}}.\end{gathered}\]

Since \(y=\sqrt{25-x^2}\) on this branch, this is exactly

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

The lower branch gives the same implicit formula once its negative value of \(y\) is substituted.

Efficiency principle

If the relation is already naturally written using both \(x\) and \(y\), implicit differentiation is usually the cleaner route. Rearranging first is useful mainly as a check, or when the explicit form is genuinely simpler.

Common mistakes and how to avoid them

Checklist before you move on

Missing the chain-rule factor: \(\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}\), not just \(ny^{n-1}\).
Forgetting the product rule: \(\dfrac{d}{dx}(xy)=x\dfrac{dy}{dx}+y\).
Using a final derivative that still has scattered \(\dfrac{dy}{dx}\)-terms: collect, factor and isolate the derivative.
Assuming the derivative must contain only \(x\): for an implicit curve, an answer in terms of both \(x\) and \(y\) is normal.
Finding a horizontal/vertical tangent from the derivative alone: also impose the original curve equation.
Taking the negative reciprocal mechanically: if the tangent is horizontal, the normal is vertical; if the tangent is vertical, the normal is horizontal.

Practice

Differentiate carefully and simplify your final result. Unless a decimal is requested, keep answers exact.

Find \(\dfrac{dy}{dx}\) if \(x^2+y^2=16\).
Find \(\dfrac{dy}{dx}\) if \(x^3+y^3=9\).
Find \(\dfrac{dy}{dx}\) if \(x^2+3xy-2y^2=10\).
Find \(\dfrac{dy}{dx}\) if \(x^2y+y^3=5\).
Find \(\dfrac{dy}{dx}\) if \(\sin y=x^2\).
Find \(\dfrac{dy}{dx}\) if \(e^y+xy=4\).
Find \(\dfrac{dy}{dx}\) if \(\ln(x+y)=x-y\).
Find \(\dfrac{dy}{dx}\) if \(\sin(x+y)=xy\).
The curve \(x^2+xy+y^2=7\) passes through \((1,2)\). Find the equations of the tangent and the normal at this point.
The ellipse \(x^2+4y^2=20\) passes through \((2,2)\). Find the equations of the tangent and the normal at this point.
For the circle \(x^2+y^2=25\), find the equations of the tangent and normal at \((3,4)\).
Find all points on \(x^2+xy+y^2=7\) where the tangent is horizontal.
Find all points on \(x^2+xy+y^2=7\) where the tangent is vertical.
For \(x^2+y^2=25\), find \(\dfrac{d^2y}{dx^2}\) at \((3,4)\).
The curve \(e^{xy}+y^2=2\) passes through \((0,1)\). Find the equations of the tangent and normal at this point.

Solution.
The curve \(x^2y+xy^2=6\) passes through \((2,1)\). Find \(\dfrac{dy}{dx}\), then find the gradient of the curve at \((2,1)\).

Answer key and solution outlines

Questions 1–4: algebraic relations

\[\begin{gathered}2x+2y\dfrac{dy}{dx}=0 \\\quad\Rightarrow\quad \boxed{\dfrac{dy}{dx}=-\frac{x}{y}}.\end{gathered}\]

\[\begin{gathered}3x^2+3y^2\dfrac{dy}{dx}=0 \\\quad\Rightarrow\quad \boxed{\dfrac{dy}{dx}=-\frac{x^2}{y^2}}.\end{gathered}\]

\[2x+3\left(y+x\dfrac{dy}{dx}\right)-4y\dfrac{dy}{dx}=0,\]

\[\begin{gathered}(3x-4y)\dfrac{dy}{dx}=-(2x+3y), \\ \boxed{\dfrac{dy}{dx}=-\frac{2x+3y}{3x-4y}}.\end{gathered}\]

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

hence

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

Questions 5–8: composite functions

\[\begin{gathered}\cos y\,\dfrac{dy}{dx}=2x \\\quad\Rightarrow\quad \boxed{\dfrac{dy}{dx}=\frac{2x}{\cos y}}.\end{gathered}\]

\[e^y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0,\]

\[\begin{gathered}(e^y+x)\dfrac{dy}{dx}=-y, \\ \boxed{\dfrac{dy}{dx}=-\frac{y}{e^y+x}}.\end{gathered}\]

\[\frac{1+\dfrac{dy}{dx}}{x+y}=1-\dfrac{dy}{dx}.\]

Multiplying by \(x+y\) and collecting derivative terms gives

\[(1+x+y)\dfrac{dy}{dx}=x+y-1,\]

therefore

\[\boxed{\dfrac{dy}{dx}=\frac{x+y-1}{x+y+1}}.\]

\[\cos(x+y)(1+\dfrac{dy}{dx})=y+x\dfrac{dy}{dx},\]

\[\boxed{\dfrac{dy}{dx}=\frac{y-\cos(x+y)}{\cos(x+y)-x}}.\]

Questions 9–11: tangents and normals

9. For \(x^2+xy+y^2=7\),

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

At \((1,2)\), \(m_t=-\dfrac45\), so

\[\boxed{4x+5y=14}.\]

The normal gradient is \(\dfrac54\), so

\[\boxed{5x-4y+3=0}.\]

10. For \(x^2+4y^2=20\),

\[\dfrac{dy}{dx}=-\frac{x}{4y}.\]

At \((2,2)\), \(m_t=-\dfrac14\). Thus

\[\boxed{x+4y=10}\]

and the normal gradient is \(4\), giving

\[\boxed{y=4x-6}.\]

11. For \(x^2+y^2=25\), \(m_t=-\dfrac34\) at \((3,4)\). Hence

\[\boxed{3x+4y=25}.\]

The normal gradient is \(\dfrac43\), so

\[\boxed{4x-3y=0}.\]

Questions 12–14: special tangents and second derivative

12. Horizontal tangent: \(2x+y=0\), so \(y=-2x\). Substitution into the curve gives \(3x^2=7\). Therefore

\[\begin{gathered}\boxed{\begin{gathered}\left(\frac{\sqrt{21}}3,-\frac{2\sqrt{21}}3\right),\\ \left(-\frac{\sqrt{21}}3,\frac{2\sqrt{21}}3\right)\end{gathered}}.\end{gathered}\]

13. Vertical tangent: \(x+2y=0\), so \(x=-2y\). Substitution gives \(3y^2=7\). Therefore

\[\begin{gathered}\boxed{\begin{gathered}\left(-\frac{2\sqrt{21}}3,\frac{\sqrt{21}}3\right),\\ \left(\frac{2\sqrt{21}}3,-\frac{\sqrt{21}}3\right)\end{gathered}}.\end{gathered}\]

14. Differentiate \(2x+2y\dfrac{dy}{dx}=0\) again:

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

At \((3,4)\), \(\dfrac{dy}{dx}=-\dfrac34\), hence

\[\boxed{\dfrac{d^2y}{dx^2}=-\frac{25}{64}}.\]

Questions 15–16: challenge

15. Differentiate

\[e^{xy}+y^2=2:\]

\[e^{xy}\left(y+x\dfrac{dy}{dx}\right)+2y\dfrac{dy}{dx}=0.\]

At \((0,1)\),

\[\begin{gathered}1+2\dfrac{dy}{dx}=0 \\\quad\Rightarrow\quad m_t=-\frac12.\end{gathered}\]

So the tangent and normal are

\[\begin{gathered}\boxed{y-1=-\frac12x}, \\ \boxed{y-1=2x}.\end{gathered}\]

16. Differentiate

\[x^2y+xy^2=6:\]

\[2xy+x^2\dfrac{dy}{dx}+y^2+2xy\dfrac{dy}{dx}=0.\]

Collecting derivative terms gives

\[(x^2+2xy)\dfrac{dy}{dx}=-(2xy+y^2),\]

\[\boxed{\dfrac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}}.\]

At \((2,1)\),

\[\boxed{\dfrac{dy}{dx}=-\frac58}.\]

Final checklist

Before finishing an implicit-differentiation question, check:

every differentiated \(y\)-expression has the required chain-rule factor \(\dfrac{dy}{dx}\);
every product such as \(xy\) has been differentiated with the product rule;
all \(\dfrac{dy}{dx}\)-terms have been collected and factored correctly;
any point used lies on the original curve;
tangent and normal gradients have been interpreted correctly;
horizontal or vertical tangent conditions have been combined with the original curve equation.