Why Nested Expressions Need Care
In implicit differentiation, \(y\) is understood to depend on \(x\), even when a relation does not give \(y\) explicitly as a function of \(x\). This matters whenever \(y\) appears inside another expression. For example, differentiating \(\sin(xy)\) requires two steps: first differentiate the inner product \(xy\), then apply the derivative rule for sine.
The same principle applies to expressions such as \((x+y)^3\) and \(e^{y^2}\). Their inner expressions depend on \(x\), partly because they contain \(y\). The Chain Rule must account for that dependence every time it occurs.
A Reliable Way to Differentiate Nested Expressions
Suppose a relation holds along a differentiable curve, so that \(y=y(x)\) and the relation remains true as \(x\) varies in an interval. The left side is then a function of \(x\). Differentiating the relation means differentiating that function of \(x\) on both sides.
For a nested expression, identify the inner expression first. If \(u=u(x)\), the Chain Rule says that the derivative of \(G(u(x))\) is \(G'(u(x))u'(x)\). In an implicit relation, \(u'(x)\) may itself contain \(y'\). For instance, with \(u=xy\), the product rule gives \(u'=y+xy'\), not merely \(y'\).
Differentiate expressions involving \(y\) as expressions in \(x\), with \(y=y(x)\).
Find the derivative of the inner expression, then multiply by the derivative of the outer function evaluated at that inner expression.
After differentiating both sides, gather terms containing \(y'\) and isolate it only when its coefficient is nonzero.
A few basic derivatives recur. The product rule gives \(\frac{d}{dx}(xy)=y+xy'\). The sum rule gives \(\frac{d}{dx}(x+y)=1+y'\). Applying the Chain Rule to a cube gives \(\frac{d}{dx}(x+y)^3=3(x+y)^2(1+y')\). Each formula includes the derivative of the entire inner expression.
Worked Example: A Sine of a Product
Worked Example: A Sine of a Product
Suppose \(y\) is differentiable and satisfies \(\sin(xy)+y^2=1\). Find an equation for \(y'\), and isolate \(y'\) when possible.
First differentiate both sides with respect to \(x\). For the first term, the outer function is sine and the inner expression is \(xy\). By the product rule, \(\frac{d}{dx}(xy)=y+xy'\). Thus
For the second term, apply the Chain Rule to \(y^2\). The derivative of its inner expression \(y\) is \(y'\), so \(\frac{d}{dx}(y^2)=2yy'\). The derivative of the constant \(1\) is \(0\). Therefore
Expand the first product and collect the terms containing \(y'\):
If \(x\cos(xy)+2y\ne 0\), division by this coefficient gives
Check by substituting this expression into the differentiated equation. Writing \(D=x\cos(xy)+2y\), the left side becomes
because the numerator of the fraction is \(y\cos(xy)D\). The differentiated relation is satisfied.
Why the Chain Rule Must Be Applied at Every Layer
The sine example illustrates the general reasoning. Along the curve, \(u(x)=x\,y(x)\) is itself a function of \(x\). The Chain Rule first gives \(\frac{d}{dx}\sin(u(x))=\cos(u(x))u'(x)\). The product rule then gives \(u'(x)=y(x)+x\,y'(x)\). Combining these equalities yields \(\cos(xy)(y+xy')\).
This is not an extra rule specific to implicit equations. It is the ordinary Chain Rule applied to a composition whose inner expression contains a dependent variable. The implicit setting makes it especially important to track the dependence rather than treating \(y\) as a constant.
Worked Example: A Power of a Nested Sum
Worked Example: A Power of a Nested Sum
Suppose \(y\) is differentiable and satisfies \((x+y)^3+x^2y=10\). Find \(y'\) wherever the resulting coefficient of \(y'\) is nonzero.
For the cubic term, the inner expression is \(x+y\), whose derivative is \(1+y'\). For the product \(x^2y\), the product rule gives \((2x)y+x^2y'\). Differentiating both sides therefore gives
Expand the cubic term and collect the \(y'\) terms:
Consequently, if \(3(x+y)^2+x^2\ne 0\), then
To check, let \(A=3(x+y)^2+x^2\) and \(B=3(x+y)^2+2xy\). The differentiated equation after collecting terms is \(Ay'+B=0\). Substitution of \(y'=-B/A\) gives \(A(-B/A)+B=-B+B=0\), as required. The substitution is valid under the stated condition \(A\ne0\).
Worked Example: An Exponential of a Dependent Expression
Worked Example: An Exponential of a Dependent Expression
Suppose \(y\) is differentiable and satisfies \(e^{y^2}+xy=4\). Find \(y'\) wherever it can be isolated.
For \(e^{y^2}\), the outer function is the exponential function and the inner expression is \(y^2\). The derivative of the inner expression is \(2yy'\), so the Chain Rule gives \(\frac{d}{dx}e^{y^2}=e^{y^2}(2yy')\). For the product \(xy\), the product rule gives \(y+xy'\). Differentiating the relation yields
Collect the terms involving \(y'\):
If \(2y e^{y^2}+x\ne0\), divide by that coefficient:
For a check, let \(D=2y e^{y^2}+x\). Substitution into the differentiated equation gives
Thus the expression satisfies the differentiated relation whenever \(D\ne0\).
A Common Mistake: Omitting the Inner Derivative
A frequent error is to differentiate an outer function correctly but omit the derivative of its inner expression. For example, the incorrect calculation
treats \(xy\) as though it did not change with \(x\). But \(xy\) depends on \(x\) both directly and through \(y=y(x)\). The correction is
Another version of the same mistake occurs with \((x+y)^3\): writing only \(3(x+y)^2\) omits the derivative \(1+y'\) of the inner expression. The correct result is \(3(x+y)^2(1+y')\). A useful check is to name the inner expression temporarily, differentiate it, and then substitute its derivative back into the Chain Rule.
When Can \(y'\) Be Isolated?
After differentiating, the terms involving \(y'\) often form a linear expression. If the resulting equation is \(A(x,y)y'=B(x,y)\), then \(y'=B(x,y)/A(x,y)\) is valid only where \(A(x,y)\ne0\). If that coefficient is zero, division is not allowed; the differentiated equation alone does not give this quotient there.
This restriction is separate from the Chain Rule. The Chain Rule gives the differentiated relation wherever the assumed differentiability permits its use. Isolating \(y'\) is an additional algebraic step with its own nonzero-denominator condition.
The central habit is to preserve every layer of dependence: differentiate the inner expression, apply the outer derivative, and repeat as needed. Once the relation has been differentiated correctly, collecting \(y'\) terms is ordinary algebra.
Check Your Understanding
For each question, assume \(y\) is differentiable and use \(y'\) to denote \(dy/dx\).
- What is \(\frac{d}{dx}\sin(xy)\), and which rule is needed to differentiate its inner expression?
- Differentiate \((x+y)^4\), showing the derivative of the inner sum.
- Differentiate \(e^{y^2}+x^2y=3\) and collect the terms containing \(y'\).
- In what circumstance is it invalid to divide by the coefficient of \(y'\) while isolating \(y'\)?
- Explain why differentiating \(\sin(xy)\) as just \(\cos(xy)\) fails to account for the dependence on \(x\).
