Implicit differentiation

Differentiate a relation without first solving it for one variable.

The idea

An explicit equation gives the output directly, such as \(y=x^2\). An implicit equation gives a relation between the variables, such as \(x^2+y^2=4\).

A relation need not define one function on its whole curve. The circle has an upper branch and a lower branch, for example. Where we can follow a differentiable branch as \(y=y(x)\), we can differentiate its equation without first solving for that branch.

The factor \(y^{\prime}\) comes from the chain rule: the variable \(y\) changes as \(x\) changes. It is not an extra symbol attached arbitrarily.

Example 1

Slopes on a circle

For \(x^2+y^2=4\), differentiate the two sides:

\[2x+2yy^{\prime}=0.\]

Solving gives

\[y^{\prime}=-\frac{x}{y},\qquad y\ne0.\]

The slope depends on both coordinates. At \((1,\sqrt3)\) it is \(-1/\sqrt3\); at \((1,-\sqrt3)\) it is \(1/\sqrt3\). The same input can lie on two branches with different slopes.

Using the derivative

Example 2

A tangent without solving a quadratic

Find the tangent to \(x^2+xy+y^2=7\) at \((1,2)\). First verify the point: \(1+2+4=7\). Differentiating gives

\[2x+y+xy^{\prime}+2yy^{\prime}=0.\]

Collect the derivative terms:

\[(x+2y)y^{\prime}=-(2x+y),\qquad y^{\prime}=-\frac{2x+y}{x+2y}.\]

At \((1,2)\), the denominator is nonzero and the slope is \(-4/5\). The tangent is

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

Example 3

A trigonometric relation

For \(\sin(y)+xy=x\), differentiate each term:

\[\cos(y)\cdot y^{\prime}+y+xy^{\prime}=1.\]

Thus

\[(\cos(y)+x)y^{\prime}=1-y,\qquad y^{\prime}=\frac{1-y}{\cos(y)+x},\qquad \cos(y)+x\ne0.\]

At \((0,0)\), the relation holds and the slope is \(1\). Notice the two different rules: the chain rule supplies \(y^{\prime}\) in the sine term, while the product rule supplies both terms from \(xy\).

Differentiate again

Example 4

A second derivative along the curve

On the circle \(x^2+y^2=4\), we already found \(x+yy^{\prime}=0\). The factor \(y^{\prime}\) is itself a function of \(x\). Differentiating the product gives

\[1+(y^{\prime})^2+yy^{\prime\prime}=0.\]

For \(y\ne0\), substitute \(y^{\prime}=-x/y\):

\[y^{\prime\prime}=-\frac{1+x^2/y^2}{y}=-\frac{x^2+y^2}{y^3}=-\frac4{y^3}.\]

At \((0,2)\), \(y^{\prime}=0\) while \(y^{\prime\prime}=-1/2\): the tangent is horizontal, but its slope changes as we move along the upper branch. At the vertical tangents, these derivatives with respect to \(x\) are not finite.

Horizontal and vertical tangents

On the circle, \(y^{\prime}=-x/y\) is zero when \(x=0\) and \(y\ne0\). Hence the top and bottom points \((0,\pm2)\) have horizontal tangents.

At \((\pm2,0)\), the tangent is vertical, so there is no finite \(dy/dx\). One way to confirm this is to regard \(x\) as a function of \(y\) instead:

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

at these points. The tangent direction has no horizontal change there, so its line is \(x=2\) or \(x=-2\).

Quiz

Enable JavaScript for interactive answer checking.A student calculates a tangent slope for \(xy=6\) at \((2,2)\). What should they check first?

Whether the point lies on the curve

Whether both coordinates are positive

Whether the curve can be solved using a square root

Explanation

Here \(2\cdot2=4\ne6\), so the point is not on the curve. Substituting coordinates into a derivative formula does not create a tangent at a point outside the curve.A rectangle keeps area \(12\,\mathrm{m^2}\). When its length \(x\) is \(3\,\mathrm{m}\) and width \(y\) is \(4\,\mathrm{m}\), what is \(dy/dx\)?

\(4/3\)

\(-4/3\)

\(-3/4\)

\(0\)

Explanation

The constraint is \(xy=12\). Differentiating gives \(y+xy^{\prime}=0\), hence \(dy/dx=-y/x=-4/3\). The width must shrink as the length grows to preserve the area.For \(x+y^2=5\), find \(dx/dy\) at \((1,2)\). Notice which variable is in the denominator.

\(-1/4\)

\(-4\)

\(4\)

\(0\)

Explanation

Differentiate with respect to \(y\): \(dx/dy+2y=0\), so \(dx/dy=-2y=-4\). The value \(-1/4\) would instead be \(dy/dx\) at this point.When \(y\) depends on \(x\), what is \(d[\cos(x+y)]/dx\)?

\(-\sin(x+y)\)

\(-\sin(x+y)\cdot y^{\prime}\)

\(-\sin(x+y)\cdot(1+y^{\prime})\)

\(\cos(x+y)\cdot(1+y^{\prime})\)

Explanation

The inner function is \(x+y(x)\), whose derivative is \(1+y^{\prime}\). The cosine rule and chain rule supply both the minus sign and this whole inner derivative.For \(y^2=x^2\), implicit differentiation gives \(y^{\prime}=x/y\) when \(y\ne0\). Does the resulting \(0/0\) at the origin prove there is a vertical tangent?

Yes: every zero denominator means a vertical tangent

No: the relation is two crossing lines, each with a finite slope

Yes: the slope must be zero

Explanation

Factor the relation as \((y-x)(y+x)=0\). It consists of \(y=x\) and \(y=-x\), with slopes \(1\) and \(-1\). The undefined quotient does not determine a tangent at the crossing.Near \((1,2)\) on \(x^3+y^3=9\), the input \(x\) increases by \(0.04\). What change in \(y\) does the tangent predict?

About \(-0.01\)

About \(0.01\)

About \(-0.16\)

Exactly \(-0.01\)

Explanation

Differentiating gives \(3x^2+3y^2y^{\prime}=0\). At \((1,2)\), the slope is \(-1/4\), so \(\Delta y\approx(-1/4)(0.04)=-0.01\). This is a local approximation, not an exact finite change.On the curve \(y^3+xy=2\), the point \((1,1)\) has slope \(y^{\prime}=-1/4\). What is \(y^{\prime\prime}\) there?

\(-1/32\)

\(1/32\)

\(0\)

\(1/4\)

Explanation

First, \((3y^2+x)y^{\prime}=-y\). Differentiate again: \((3y^2+x)y^{\prime\prime}+(6yy^{\prime}+1)y^{\prime}=-y^{\prime}\). At the point this gives \(4y^{\prime\prime}=1/2-6/16=1/8\), hence \(y^{\prime\prime}=1/32\).