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Ify 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

`(dy)/(dx)=(dy)/(du)*(du)/(dx)`

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#### Solution

Let δx be a small increment in x.

Let δy and δu be the corresponding increments in y and u respectively

As δx → 0, δy → 0, δu → 0.

As u is differentiable function, it is continuous.

Consider the incrementary ratio `(deltay)/(deltax)`

`"We have ",(deltay)/(deltax)=(deltay)/(deltau)xx(deltau)/(deltax)`

Taking limit as δx → 0, on both sides,

`lim_(deltax->0)(deltay)/(deltax)=lim_(deltax->0)((delty)/(deltau)xx(deltau)/(deltax))`

`lim_(deltax->0)(deltay)/(deltax)=lim_(deltau->0)(deltay)/(deltau)xxlim_(deltax->0)(deltau)/(deltax)...(1)`

Since y is a differentiable function of u , `lim_(deltau->0)(deltay)/(deltau)` exists

and `lim_(deltau->0)(deltay)/(deltax) ` exists as u is a differentiable function of x.

Hence, R.H.S. of (1) exists

`"now " lim_(deltau->0)(deltay)/(deltau)=(dy)/(du) and lim_(deltau->0)(deltau)/(deltax)=(du)/(dx)`

`lim_(deltax->0)(deltay)/(deltax)=(dy)/(du)xx(du)/(dx)`

Since R.H.S. exists, L.H.S. of (1) also exists and

`lim_(deltax->0)(deltay)/(deltax)=(dy)/(dx)`

`dy/dx=(dy)/(du)xx(du)/(dx)`