Question

# If $y=f\left[ \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right]$ , then $\dfrac{dy}{dx}$ is equal to(a) ${f}'\left( \phi o\psi ohog \right)\centerdot {\phi }'\left( \psi ohog \right)\centerdot {h}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)$ (b) $\dfrac{dy}{dx}\centerdot \dfrac{d\phi }{dh}$ (c) $\dfrac{d\phi }{dh}\centerdot \dfrac{dh}{dx}$ (d) none of these

Hint: First, we can see clearly that the function here mentioned is a composite function which is the combination of several functions. Then, we need to use the chain rule to find the derivative of such types of functions. Then, we can clearly see that by chain rule the continuous differentiation of the function continues till it reaches the value of differentiation of x which is 1 to get the desired answer.

In this question, we are supposed to find the differentiation of the function $y=f\left[ \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right]$ .
Now, we can see clearly that the function here mentioned is a composite function which is the combination of several functions.
So, we need to use the chain rule to find the derivative of such type of functions.
Now, before proceeding for the actual function let the function b y=f(g(x)) and its differentiation by chain rule is given by:
\begin{align} & \dfrac{dy}{dx}={f}'\left( g\left( x \right) \right)\centerdot \dfrac{d}{dx}g\left( x \right) \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)\centerdot \dfrac{d}{dx}x \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)\centerdot 1 \\ \end{align}
So, we can clearly that by chain rule the continuous differentiation of the function continues till it reaches the value of differentiation of x which is 1.
Now, by applying the same rule to the given function in the question as $y=f\left[ \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right]$ and solving it by chain rule, we get:
\begin{align} & \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot \dfrac{d}{dx}\left[ \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right] \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\dfrac{d}{dx}\left[ \psi \left( h\left( g\left( x \right) \right) \right) \right] \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\centerdot {\psi }'\left( h\left( g\left( x \right) \right) \right)\dfrac{d}{dx}\left[ h\left( g\left( x \right) \right) \right] \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\centerdot {\psi }'\left( h\left( g\left( x \right) \right) \right)\centerdot {h}'\left( g\left( x \right) \right)\dfrac{d}{dx}\left[ g\left( x \right) \right] \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\centerdot {\psi }'\left( h\left( g\left( x \right) \right) \right)\centerdot {h}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)\centerdot \dfrac{d}{dx}x \\ & \Rightarrow \dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\centerdot {\psi }'\left( h\left( g\left( x \right) \right) \right)\centerdot {h}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right) \\ \end{align}
So, the differentiation of the function $y=f\left[ \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right]$ is ${f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)\centerdot {\phi }'\left( \psi \left( h\left( g\left( x \right) \right) \right) \right)\centerdot {\psi }'\left( h\left( g\left( x \right) \right) \right)\centerdot {h}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)$ .
Now, the answer above found can be written as ${f}'\left( \phi o\psi ohog \right)\centerdot {\phi }'\left( \psi ohog \right)\centerdot {h}'\left( g\left( x \right) \right)\centerdot {g}'\left( x \right)$ .
So, the correct answer is “Option A”.

Note: Now, the only mistake we can make in these types of questions is that we don’t apply the chain rule and just solve till the first bracket and give that as an answer. So, if we do the same with this question our answer will be as:
$\dfrac{dy}{dx}={f}'\left( \phi \left( \psi \left( h\left( g\left( x \right) \right) \right) \right) \right)$ And we give it an answer which doesn’t match with any option and we will mark one of these answer which is the wrong option. So, be careful while performing differentiation of the composite functions.