Question

Differentiate the following w.r.t. x: (sin x^x)

Answer 1

Let y = (sin x^x)

Then `"dy"/"dx" = "d"/"dx"[(sinx^x)]`

∴ `"dy"/"dx" = cos(x^x)."d"/"dx"(x^x)`          ...(1)
Let u  = x^x

Then log u = logx^x = x.logx

Differentiating both sides w.r.t. x, we get

`1/u."du"/"dx" = "d"/"dx"(x.logx)`

= `x."d"/"dx"(logx) + (logx)."d"/"dx"(x)`

= `x xx (1)/x + (logx) xx 1`

∴ `"du"/"dx" = u(1 + logx)`

∴ `"d"/"dx"(x^x) = x^x (1 + logx)`             ...(2)

From (1) and (2), we get

`"dy"/"dx" = cos(x^x).x^x(1 + logx)`