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Proof of Chain rule in Leibniz’s notation

The chain rule is derived fundamentally to find the differentiation of any composite function $f\Big(g(x)\Big)$.

$\dfrac{d}{dx}\,f\Big(g(x)\Big)$ $\,=\,$ $\dfrac{d}{d\,g(x)}\,f\Big(g(x)\Big) \times \dfrac{d}{dx}\,g(x)$

This chain rule is simply written in Leibnitz’s form as follows.

$\dfrac{dy}{dx} \,=\, \dfrac{dy}{dz} \times \dfrac{dz}{dx}$

Now, let’s learn how to prove the chain rule in Leibniz’s notation mathematically in calculus.

Differentiate the Composite function

Assume $y \,=\, f\Big(g(x)\Big)$

Now, differentiate the expressions on both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}\,y \,=\, \dfrac{d}{dx}\,f\Big(g(x)\Big)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}\,f\Big(g(x)\Big)$

Differentiate the Internal function

Suppose $z \,=\, g(x)$

Now, differentiate the both expressions on both sides of the equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}\,z \,=\, \dfrac{d}{dx}\,g(x)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}\,g(x)$

Relation between variables in differentials

We have assumed that $z = g(x)$. Hence, $y \,=\, f\Big(g(x)\Big)$ can be written as $y \,=\, f(z)$. Now, differentiate the equation with respect to $z$.

$\implies$ $\dfrac{d}{dz}\,y \,=\, \dfrac{d}{dz}\,f(z)$

$\implies$ $\dfrac{dy}{dz} \,=\, \dfrac{d}{dz}\,f(z)$

We know that $z = g(x)$. So, replace the variable $z$ by its actual value on the right-hand side of the equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dz} \,=\, \dfrac{d}{d\,g(x)}\,f\Big(g(x)\Big)$

Proving the Chain Rule in Leibniz’s form

It is time to remind all the results one by one to prove the chain Rule in Leibnitz’s notation.

$(1).\,\,\,$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}\,f\Big(g(x)\Big)$

$(2).\,\,\,$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}\,g(x)$

$(3).\,\,\,$ $\dfrac{dy}{dz} \,=\, \dfrac{d}{d\,g(x)}\,f\Big(g(x)\Big)$

Look at the following mathematical steps.

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{dy}{dx} \times 1$

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{dy}{dx} \times \dfrac{dz}{dz}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx} \,=\, \dfrac{dy}{dz} \times \dfrac{dz}{dx}$

Mathematically, the derivative of $y$ with respect to $x$ can be obtained by multiplying the derivatives of $y$ and $z$ with respect to $z$ and $x$ respectively.

$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{dy}{dz} \times \dfrac{dz}{dx}$

Now, substitute the above differentials by their respective values to get the chain rule mathematically.

$\implies$ $\dfrac{d}{dx}\,f\Big(g(x)\Big)$ $\,=\,$ $\dfrac{d}{d\,g(x)}\,f\Big(g(x)\Big) \times \dfrac{d}{dx}\,g(x)$

This mathematical equation in differential form expresses the method to find the derivative of composition of two or more functions. It is called the chain rule. Hence, the following mathematical equation is other form of chain rule in calculus.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx} \,=\, \dfrac{dy}{dz} \times \dfrac{dz}{dx}$

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