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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
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Solution
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
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