Implicit and Logarithmic Differentiation

A curve need not arrive as y equals a formula

An equation $F(x,y)=0$ can describe a curve without explicitly solving for $y$ . Along a differentiable branch $y(x)$ , every occurrence of $y$ depends on $x$ . Applying the chain rule to $y^2$ therefore gives $2yy'$ , not $2y$ .

Differentiate both sides, collect all terms containing $y'$ , and solve where their coefficient is nonzero. The resulting formula can fail at a point even when the curve is geometrically well behaved; the curve may have a vertical tangent or may not define one function near that point.

Visual guide

VISUAL GUIDEDifferentiate along a level curve

Unit circle
Tangent at the marked point

On x² + y² = 1, implicit differentiation gives y′ = −x/y when y ≠ 0. At (√½, √½), the tangent slope is −1. At y = 0 the graph has a vertical tangent, so this formula cannot give a finite slope.

Worked example: a circle

For $x^2+y^2=25$ , differentiation gives $2x+2yy'=0$ . If $y\ne0$ ,

y'=-x/y.

At $(3,4)$ the tangent slope is $-3/4$ , so the line is $y-4=-(3/4)(x-3)$ . At $(5,0)$ this formula cannot supply a finite slope. The circle has a vertical tangent there, and solving for $x$ as a function of $y$ is more suitable locally.

For a product such as $xy$ , both factors contribute: its derivative is $y+xy'$ . Skipping the product rule is a common source of an incorrect implicit derivative.

Worked example: a variable exponent

For $y=x^x$ with $x>0$ , neither the ordinary power rule nor the constant-base exponential rule applies directly: both base and exponent vary. Taking logarithms gives $\ln y=x\ln x$ . Differentiate:

y'/y=\ln x+1,

y'=x^x(\ln x+1).

The positive-domain restriction made the logarithms legitimate. For products and quotients of nonzero factors on an interval, using logarithms of absolute values can turn products into sums while retaining derivative identity $(\ln|u|)'=u'/u$ .

Inverse functions as implicit relations

If $y=f^{-1}(x)$ , then $f(y)=x$ . Differentiating gives $f'(y)y'=1$ , hence $(f^{-1})'(x)=1/f'(f^{-1}(x))$ where the denominator is nonzero and the inverse branch is differentiable. For $y=\arcsin x$ , $\sin y=x$ with $y\in[-\pi/2,\pi/2]$ . On $-1<x<1$ , $y'=1/\sqrt{1-x^2}$ because cosine is positive on the interior of that branch.

Practice

Find $y'$ for $xy+y^2=6$ where $x+2y\ne0$ .
Differentiate $x^{2x}$ for $x>0$ .
Why does the inverse derivative formula fail for $f(x)=x^3$ at zero?

Show worked solutions

$y+xy'+2yy'=0$ , so $y'=-y/(x+2y)$ .
$\ln y=2x\ln x$ gives $y'=x^{2x}(2\ln x+2)$ .
$f'(0)=0$ , so the reciprocal formula has a zero denominator. The inverse cube-root function has no finite derivative at zero.