Tag: Chain Rule

When a function is built by plugging one function into another, we call it a composite function. The Chain Rule is the engine that differentiates such compositions. This article gives intuition, the formal statement, a clean proof, and worked examples you can copy for practice. Intuition: Rates Inside Rates If a quantity $latex y$ depends on an intermediate variable $latex u$, and $latex u$ depends on $latex x$, then changes in $latex x$ ripple through $latex u$ to affect $latex y$. Symbolically, let $latex y=f(u)$ and $latex u=g(x)$. A small change $latex \Delta x$ produces $latex \Delta u \approx g'(x)\,\Delta x$, which then produces $latex \Delta y \approx f'(u)\,\Delta u$. Multiplying the rates suggests $latex \displaystyle \frac{dy}{dx} \approx f'(u)\,g'(x)$, evaluated at $latex u=g(x)$. The Chain Rule makes this exact. Formal Statement (Two Equivalent Forms) Function Composition Form If $latex g$ is differentiable at $latex x$ and $latex f$ is differentiable at $latex g(x)$, then the composite $latex F(x)=f(g(x))$ is differentiable at $latex x$, and $latex F'(x)=f'\!\big(g(x)\big)\cdot g'(x).$ Leibniz Form Writing $latex y=f(u)$ and $latex u=g(x)$ with both differentiable, $latex \displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx},$ where $latex \displaystyle\frac{dy}{du}$ is evaluated at $latex u=g(x)$ and $latex \displaystyle\frac{du}{dx}$ at $latex x$. A Proof of the Chain Rule Setup Let $latex y=f(u)$ with $latex f$ differentiable at $latex b=g(a)$, and let $latex u=g(x)$ be differentiable at $latex a$. Define the increments $latex \Delta x$, $latex \Delta u=g(a+\Delta x)-g(a)$, and $latex \Delta y=f(b+\Delta u)-f(b)$. By the definition of differentiability, there exist functions $latex \varepsilon_1(\Delta x)$ and $latex \varepsilon_2(\Delta u)$ with $latex \varepsilon_1\to 0$ as $latex \Delta x\to 0$ and $latex \varepsilon_2\to 0$ as $latex \Delta u\to 0$ such that $latex \Delta u = g'(a)\,\Delta x + \varepsilon_1(\Delta x)\,\Delta x = \big[g'(a)+\varepsilon_1(\Delta x)\big]\Delta x,$ $latex \Delta y = f'(b)\,\Delta u + \varepsilon_2(\Delta u)\,\Delta u = \big[f'(b)+\varepsilon_2(\Delta u)\big]\Delta u.$ Combine Substitute the expression for $latex \Delta u$ into $latex \Delta y$: $latex \Delta y = \big[f'(b)+\varepsilon_2(\Delta u)\big]\big[g'(a)+\varepsilon_1(\Delta x)\big]\Delta x.$ Divide by $latex \Delta x$: $latex \displaystyle \frac{\Delta y}{\Delta x} = \big[f'(b)+\varepsilon_2(\Delta u)\big]\big[g'(a)+\varepsilon_1(\Delta x)\big].$ Take Limits As $latex \Delta x\to 0$, we have $latex \Delta u\to 0$, so $latex \varepsilon_1(\Delta x)\to 0$ and $latex \varepsilon_2(\Delta u)\to 0$. Therefore, $latex \displaystyle \frac{dy}{dx} = \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} = f'(b)\,g'(a) = f'\!\big(g(a)\big)\,g'(a).$ Since $latex a$ was arbitrary, the formula holds for every $latex x$ in the domain. This proves the Chain Rule. Useful Special Case: Power Rule + Chain Rule If $latex n$ is any real number and $latex u=g(x)$ is differentiable, then $latex \displaystyle \frac{d}{dx}\big(u^n\big) = n\,u^{\,n-1}\,\frac{du}{dx} = n\,[g(x)]^{\,n-1}\,g'(x).$ Worked Examples Example 1: A Root of a Quadratic Let $latex F(x)=\sqrt{x^2+3}$. Write $latex u=x^2+3$, so $latex F(x)=u^{1/2}$. Then $latex F'(x)=\displaystyle \frac{1}{2}\,u^{-1/2}\cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}}\cdot 2x = \displaystyle \frac{x}{\sqrt{x^2+3}}.$ Created with Desmos Example 2: Sine of a Square Differentiate $latex y=\sin(x^2-5)$. Take $latex u=x^2-5$. $latex \displaystyle \frac{dy}{dx}=\cos(u)\cdot \frac{du}{dx} =\cos(x^2-5)\cdot 2x =2x\cos(x^2-5).$ Created with Desmos Example 3: Squared Sine Differentiate $latex y=\big(\sin 2x\big)^2$. View it as outer $latex (\cdot)^2$ and inner $latex \sin(2x)$: $latex \displaystyle \frac{dy}{dx} =2\sin(2x)\cdot \frac{d}{dx}[\sin(2x)] =2\sin(2x)\cdot \cos(2x)\cdot 2 =4\sin(2x)\cos(2x).$ (Equivalently, $latex y' = 2\sin(2x)\cos(2x)=\sin(4x)$ by a trig identity.) Created with Desmos Example 4: A Quotient Inside a Power Let $latex g(t)=\left(\displaystyle\frac{t+1}{3t-2}\right)^{7}$. Then $latex g'(t) =7\left(\displaystyle\frac{t+1}{3t-2}\right)^{6} \cdot \displaystyle\frac{d}{dt}\!\left(\displaystyle\frac{t+1}{3t-2}\right) =7\left(\displaystyle\frac{t+1}{3t-2}\right)^{6} \cdot \displaystyle\frac{(3t-2)\cdot 1-(t+1)\cdot 3}{(3t-2)^2}$ $latex \quad =7\left(\displaystyle\frac{t+1}{3t-2}\right)^{6} \cdot \displaystyle\frac{3t-2-3t-3}{(3t-2)^2} = -\,\displaystyle\frac{35}{(3t-2)^{\,8}}\,(t+1)^{6}.$ Created with Desmos Example 5: Product + Chain Differentiate $latex y=(x-1)^{4}\,(x^3+2)^{5}$. Use the Product Rule and then the Chain Rule: $latex \displaystyle \frac{dy}{dx} =(x-1)^{4}\cdot 5(x^3+2)^{4}\cdot 3x^{2} +(x^3+2)^{5}\cdot 4(x-1)^{3}.$ You may factor $latex (x-1)^{3}(x^3+2)^{4}$ for a cleaner final form if desired. Created with Desmos Example 6: Exponential of a Trig Function Differentiate $latex y=e^{\cos(3x)}$. $latex \displaystyle \frac{dy}{dx} =e^{\cos(3x)}\cdot \frac{d}{dx}[\cos(3x)] =e^{\cos(3x)}\cdot (-\sin(3x))\cdot 3 =-3e^{\cos(3x)}\sin(3x).$ Created with Desmos Example 7: Two Links in the Chain If $latex f(x)=\sin(\cos(x^2))$, then $latex f'(x)=\cos(\cos(x^2))\cdot \frac{d}{dx}[\cos(x^2)] =\cos(\cos(x^2))\cdot\big(-\sin(x^2)\big)\cdot 2x =-2x\,\cos(\cos(x^2))\,\sin(x^2).$ Created with Desmos Common Pitfalls Forgetting to multiply by the inner derivative. Every outer differentiation must be followed by the derivative of the inside. Evaluating at the wrong place. In $latex f'(g(x))$, the outer derivative is computed at $latex u=g(x)$, not at $latex x$. Product vs. chain. If factors are multiplied (not composed), apply the Product Rule first, then the Chain Rule to each factor as needed. Takeaway The Chain Rule turns complicated compositions into manageable steps: differentiate the outer, keep the inside unchanged, and multiply by the derivative of the inner. Mastering this pattern unlocks derivatives across algebraic, trigonometric, exponential, and logarithmic families. See more in: Wikipedia, Khan Academy, and MIT OCW.