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# If y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x, then prove that y = f (g (x) ) is a differentiable function of x and $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ .  Hint:We have been given that y = f (u) and u = g (x), and also we are given that, y = f (g (x) ). We will differentiate y = f (g (x) ) using the chain rule, and also we will differentiate y = f (u) and u = g (x), and then solving these further we will get our answer.

It is given in the question that y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x, then we have been asked to prove that y = f (g (x) ) is a differentiable function of x and $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ .
So, we have y = f (u) and u = g (x) and also it is given that y = f (g (x) ).
On differentiating y with respect to x, we get,
$\dfrac{dy}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right).........\left( i \right)$
On differentiating f with respect to u, we get,
$\dfrac{dy}{du}=f'\left( u \right).........\left( ii \right)$
On differentiating u with respect to x, we get,
$\dfrac{du}{dx}=g'\left( x \right).........\left( iii \right)$
Now, if we put g’ (x) as $\dfrac{du}{dx}$ in equation (i), we will get,
$\dfrac{dy}{dx}=f'\left( g\left( x \right) \right).\dfrac{du}{dx}$
Now, we know that g (x) = u, so we get,
$\dfrac{dy}{dx}=f'\left( u \right).\dfrac{du}{dx}.........\left( iv \right)$
Now, from equation (ii), we know that $\dfrac{dy}{du}=f'\left( u \right)$ . So, on putting the value of f’ (u) as $\dfrac{dy}{du}$ in equation (iv), we get,
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$
Hence proved.

Note: Most of the students make mistakes while differentiating f (g (x) ), they differentiate it as f’ (g (x) ) and skip to differentiate g (x), which violates the chain rule and leads to the formation of wrong answers. So, it is recommended to do the differentiation properly.
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