The Chain Rule when y Depends on x

Why Does $y^2$ Produce a $dy/dx$ Factor?

When $y$ depends on $x$, an expression such as $y^2$ changes as $x$ changes. Differentiating $y^2$ with respect to $x$ must therefore account for two things: how the squaring operation changes its input, and how $y$ changes with $x$.

The chain rule combines these two rates of change. In this setting, it gives

\frac{d}{dx}(y^2)=2y\frac{dy}{dx}.

The factor $2y$ is the derivative of the outer function, the square. The factor $dy/dx$ records the rate at which the inner quantity $y$ changes as $x$ changes. Omitting that second factor would treat $y$ as though it were simply the independent variable.

Key idea: If $y=y(x)$ is differentiable, then differentiating an expression in $y$ with respect to $x$ requires the chain rule. The derivative of the outer expression is multiplied by $dy/dx$.

What the Derivative Notation Means

Suppose $y$ is a differentiable function of $x$, so $y=y(x)$. The notation $dy/dx$ means the derivative of that function with respect to $x$. Thus, $\frac{d}{dx}(y^2)$ is the derivative with respect to $x$ of the composite function $y(x)^2$.

It is useful to temporarily give the outer function its own input name. Let $g(u)=u^2$. Then $y(x)^2=g(y(x))$, a composition of $g$ with $y$. The chain rule says that the derivative of this composition is the derivative of $g$ evaluated at $y(x)$, multiplied by $y'(x)$.

\frac{d}{dx}g(y(x))=g'(y(x))y'(x).

Since $g'(u)=2u$, substituting $u=y(x)$ gives $g'(y(x))=2y(x)$. Therefore,

\frac{d}{dx}(y^2)=2y\frac{dy}{dx}.

The notation $2y\frac{dy}{dx}$ does not mean that $y$ is an independent variable. It is a compact way to show that the derivative of the square, $2y$, is multiplied by the derivative of $y$ with respect to $x$.

A Derivation Using the Difference Quotient

The formula for $y^2$ can also be justified directly from the definition of a derivative. This makes clear why the factor $dy/dx$ is necessary.

Start with the difference quotient.
For $y=y(x)$, the derivative of $y(x)^2$ at $x$ is the limit of $\big(y(x+h)^2-y(x)^2\big)/h$, if that limit exists.

Factor the difference of squares.
Write the numerator as $\big(y(x+h)-y(x)\big)\big(y(x+h)+y(x)\big)$.

Separate the two changing factors.
For $h\ne 0$, the quotient is $\big(y(x+h)-y(x)\big)/h$ multiplied by $y(x+h)+y(x)$.

Take the limit.
Differentiability of $y$ implies continuity of $y$, so the first factor tends to $y'(x)$, while the second tends to $2y(x)$.

\begin{aligned} \frac{d}{dx}\big(y(x)^2\big) &=\lim_{h\to 0}\frac{y(x+h)^2-y(x)^2}{h}\\ &=\lim_{h\to 0} \left(\frac{y(x+h)-y(x)}{h}\right) \big(y(x+h)+y(x)\big)\\ &=y'(x)\big(2y(x)\big)\\ &=2y\frac{dy}{dx}. \end{aligned}

Both limits used in the final step exist: the first because $y$ is differentiable at $x$, the second because differentiability implies continuity. This argument proves the formula at every point where $y$ is differentiable.

Positive Integer Powers of $y$

The same reasoning applies to any positive integer power. Treat the power as the outer function $g(u)=u^n$, where $n$ is a positive integer. Its derivative is $g'(u)=nu^{n-1}$. Applying the chain rule to $g(y(x))$ gives the following result.

Theorem: If $y=y(x)$ is differentiable and $n$ is a positive integer, then $$ \frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}. $$ The derivative of the outer power gives $ny^{n-1}$; the derivative of the inner function gives $dy/dx$.

For $n=2$, the result is $2y\,dy/dx$. For $n=3$, it is $3y^2\,dy/dx$. In each case, the derivative is with respect to $x$, even though the power is written using $y$.

Worked Example: Differentiate a Square When the Rate of Change Is Given

Suppose $y$ is differentiable, $y(2)=3$, and $y'(2)=-4$. Find $\frac{d}{dx}(y^2)$ at $x=2$.

First apply the square rule:

\frac{d}{dx}(y^2)=2y\frac{dy}{dx}.

Now use the given values $y(2)=3$ and $y'(2)=-4$:

\left.\frac{d}{dx}(y^2)\right|_{x=2} =2y(2)y'(2) =2(3)(-4) =-24.

Check: the square contributes the factor $2y(2)=6$, and the changing value of $y$ contributes $y'(2)=-4$. Their product is $6(-4)=-24$, matching the chain-rule result. The negative sign indicates that $y^2$ is decreasing at $x=2$ at that instant.

Worked Example: Differentiate a Combination of Powers of $y$

Suppose $y=y(x)$ is differentiable. Differentiate $y^5-2y^2$ with respect to $x$.

Differentiate each power using the chain rule. For the first term, the outer derivative is $5y^4$, and the inner derivative is $dy/dx$:

\frac{d}{dx}(y^5)=5y^4\frac{dy}{dx}.

For the second term, keep the constant multiplier $-2$ and apply the square rule:

\frac{d}{dx}(-2y^2) =-2\left(2y\frac{dy}{dx}\right) =-4y\frac{dy}{dx}.

Combining the derivatives gives

\frac{d}{dx}(y^5-2y^2) =5y^4\frac{dy}{dx}-4y\frac{dy}{dx}.

A common factor may be taken out, but is not required:

\frac{d}{dx}(y^5-2y^2) =(5y^4-4y)\frac{dy}{dx}.

Check: distributing the factored expression gives $5y^4\,dy/dx-4y\,dy/dx$, exactly the two term-by-term derivatives above.

Sine of a Function

The same chain rule applies when the outer function is trigonometric. Let $g(u)=\sin u$, using radians. Then $g'(u)=\cos u$. For differentiable $y=y(x)$, the composition is $\sin(y(x))$, so

Chain-rule formula: If $y=y(x)$ is differentiable, then $$ \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}. $$ The cosine factor comes from differentiating sine with respect to its input; the $dy/dx$ factor comes from differentiating that input with respect to $x$.

The formula is not $\cos y$ alone unless $dy/dx=1$. Sine changes in response to changes in its input, and the input $y$ may itself change faster or slower than $x$.

Worked Example: Differentiate Sine of a Linear Function

Let $y(x)=\frac{\pi}{3}+2x$. Find $\frac{d}{dx}(\sin y)$ at $x=0$.

First find the inner value and its derivative:

y(0)=\frac{\pi}{3}+2(0)=\frac{\pi}{3}, \qquad \frac{dy}{dx}=2.

Apply the sine chain rule and substitute these values:

\left.\frac{d}{dx}(\sin y)\right|_{x=0} =\cos(y(0))\,y'(0) =\cos\left(\frac{\pi}{3}\right)(2) =\frac{1}{2}(2) =1.

Check by writing the composite function explicitly: $\sin y=\sin\left(\frac{\pi}{3}+2x\right)$. The chain rule for this composition gives $2\cos\left(\frac{\pi}{3}+2x\right)$, which at $x=0$ is $2\cdot\frac12=1$, the same result.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to differentiate the outer expression correctly but stop before multiplying by $dy/dx$. For instance, the incorrect calculation

\frac{d}{dx}(y^2)=2y

treats $y$ as though it were the independent variable. But the requested derivative is with respect to $x$, and $y$ depends on $x$. The corrected calculation is

\frac{d}{dx}(y^2)=2y\frac{dy}{dx}.

For a concrete check, take $y(x)=3x$. Then $y^2=9x^2$, whose derivative is $18x$. The chain-rule expression gives $2y\,dy/dx=2(3x)(3)=18x$. The incorrect expression $2y=6x$ does not match the derivative of $9x^2$.

Check before finishing: When differentiating with respect to $x$, ask whether the quantity inside the outer function depends on $x$. If it does, include its derivative as the inner factor.

Why This Rule Matters

The notation $y$ can make it easy to overlook that it represents a changing function rather than a fixed value. The chain rule makes that dependence explicit. It lets us differentiate powers, trigonometric expressions, and other functions of $y$ while keeping track of how $y$ changes with $x$.

The pattern is consistent: identify the outer function, differentiate it with respect to its input, evaluate the result at $y$, and multiply by $dy/dx$. For example, a power uses the derivative of $u^n$, while sine uses the derivative of $\sin u$. In both cases the inner derivative is essential.

Check Your Understanding

For each question, remember that $y$ is a differentiable function of $x$.

Write the derivative of $y^2$ with respect to $x$, including the inner derivative factor.
Write the general chain-rule formula for the derivative of $y^n$, where $n$ is a positive integer.
If $y(1)=2$ and $y'(1)=5$, what is the value of $\frac{d}{dx}(y^2)$ at $x=1$?
Differentiate $\sin y$ with respect to $x$.
For $y(x)=\frac{\pi}{6}+3x$, find $\frac{d}{dx}(\sin y)$ at $x=0$.

The Chain Rule when y Depends on x

What You'll Learn

Why Does \(y^2\) Produce a \(dy/dx\) Factor?

What the Derivative Notation Means

A Derivation Using the Difference Quotient

Positive Integer Powers of \(y\)

Worked Example: Differentiate a Square When the Rate of Change Is Given

Worked Example: Differentiate a Combination of Powers of \(y\)

Sine of a Function

Worked Example: Differentiate Sine of a Linear Function

A Common Mistake: Forgetting the Inner Derivative

Why This Rule Matters

Check Your Understanding

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

The Chain Rule when y Depends on x

What You'll Learn

Why Does \(y^2\) Produce a \(dy/dx\) Factor?

What the Derivative Notation Means

A Derivation Using the Difference Quotient

Positive Integer Powers of \(y\)

Worked Example: Differentiate a Square When the Rate of Change Is Given

Worked Example: Differentiate a Combination of Powers of \(y\)

Sine of a Function

Worked Example: Differentiate Sine of a Linear Function

A Common Mistake: Forgetting the Inner Derivative

Why This Rule Matters

Check Your Understanding

Test yourself: AP Calculus practice exams

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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