# 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 - Mathematics and Statistics

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)

#### 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)

Is there an error in this question or solution?
2015-2016 (March)

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