Math Doubts

Proof of Chain Rule

Formula

$\dfrac{d}{dx} {f[{g(x)}]} \,=\, {f'[{g(x)}]}.{g'{(x)}}$

Let $f(x)$ and $g(x)$ be two functions in terms of $x$ and their composition formed a composite function, denoted by $f\Big(g(x)\Big)$. The derivative of $f\Big(g(x)\Big)$ with respect to $x$ is written mathematically as follows in calculus.

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

Proof

$f{(x)}$ is a function and $g{(x)}$ is another function. $f{(g{(x)})}$ is a composition of the both functions.

Representation of Derivatives of the functions

The differentiation of a function is represented in short form in calculus as follows. It is used in deriving the chain rule. So, remember them.

$(1) \,\,\,\,\,\,$ $\dfrac{d}{dx} f{(x)} \,=\, f'{(x)}$

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

The derivative of function $f{[g{(x)}]}$ with respect to $x$ is written mathematically as follows.

$\dfrac{d}{dx}{f{[g{(x)}]}}$

Transform the function in known form

Generally, $f{(x)}$ is a known form of the function. So, try to convert the function $f{[g{(x)}]}$ to the known form for simplifying the complexity of the function.

Take $y = g{(x)}$. Therefore, the function $f{[g{(x)}]}$ can be simplify written as $f{(y)}$.

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

Change the differential element

The function $f{(y)}$ is in terms of $y$ but the differential element is in terms of $x$. It is not possible to differentiate the function $f{(y)}$ with respect to $x$. Therefore, convert the differential element ($dx$) from $x$ to $y$.

It is possible by differentiating $y = g{(x)}$ with respect to $x$.

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

$\implies \dfrac{dy}{dx} \,=\, g'{(x)}$

$\implies \dfrac{dy}{g'{(x)}} \,=\, dx$

$\,\,\, \therefore \,\,\,\,\,\, dx \,=\, \dfrac{dy}{g'{(x)}}$

Now, replace the differential element from $x$ to $y$ term.

$\implies \dfrac{d}{dx}{f{[g{(x)}]}}$ $\,=\,$ $\dfrac{d}{\dfrac{dy}{g'{(x)}}}{f{(y)}}$

The function $g'{(x)}$ divides the differential element $dy$. So, it multiplies $d$ in numerator.

$\implies \dfrac{d}{dx}{f{[g{(x)}]}}$ $\,=\,$ $\dfrac{g'{(x)} \times d}{dy}{f{(y)}}$

$\implies \dfrac{d}{dx}{f{[g{(x)}]}}$ $\,=\,$ $g'{(x)} \dfrac{d}{dy}{f{(y)}}$

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

Get the Chain Rule

Now, replace the value of $y$. Actually, it is $y = g{(x)}$. Therefore, replace $y$ by its value in the composition of the functions.

$\implies \dfrac{d}{dx}{f{[g{(x)}]}}$ $\,=\,$ $g'{(x)} f'{[{g{(x)}}]}$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx}{f{[g{(x)}]}}$ $\,=\,$ $f'{[{g{(x)}}]}g'{(x)}$

This property is called as chain rule in differential calculus and it is used as a formula while dealing the functions which are formed compositions of two or more functions.

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