Question

If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x such that the composite function y = f[g(x)] is a differentiable function of x, then `("d"y)/("d"x) = ("d"y)/("d"u)*("d"u)/("d"x)`. Hence find `("d"y)/("d"x)` if y = sin²x

Answer 1

Let δx be a small increment in the value of x.

Since u is a function of x, there should be a corresponding increment δu in the value of u.

Also y is a function of u.

∴ There should be a corresponding increment δy in the value of y.

Consider, `(deltay)/(deltax) = (deltay)/(deltau) xx (deltau)/(deltax)`

Taking `lim_(deltax -> 0)` on both sides, we get

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

As δx → 0, δu → 0  ........[u is a continuous function of x]

∴ `lim_(deltax -> 0) (deltay)/(deltax) = lim_(deltau -> 0) (deltay)/(deltau) xx lim_(deltax ->0) (deltau)/(deltax)`  ........(i)

y is a differentiable function of u and u is a differentiable function of x.

∴ `lim_(deltau -> 0) (deltay)/(deltau) = ("d"y)/("d"u)` exists and is finite.

Also, `lim_(deltax -> 0) (deltau)/(deltax) = ("d"u)/("d"x)` exists and is finite.

From (i), we get

`lim_(deltax -> 0) (deltay)/(deltax) = ("d"y)/("d"u) xx ("d"u)/("d"x)`  ........(ii)

Here, R.H.S. of (ii) exists and is finite.

Hence, L.H.S. of (ii) should also exists and be finite.

∴ `lim_(deltax -> 0) (deltay)/(deltax) = ("d"y)/("d"x)` exists and is finite.

∴ Equation (ii) becomes

`("d"y)/("d"x) = ("d"y)/("d"u) xx ("d"u)/("d"x)`

y = sin²x

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "d"/("d"x)(sin^2x)`

= `2sinx*"d"/("d"x)(sinx)`

= 2 sin x cos x