Question

Differentiate the following w.r.t. x : `(x^2 + 3)^(3/2).sin^3 2x.2^(x^2)`

Answer 1

Let y = `(x^2 + 3)^(3/2).sin^3 2x.2^(x^2)`

Then log y = `log[x^2 + 3)^(3/2).sin^3 2x.2^(x^2)`

= `log(x^2 + 3)^(3/2) + logsin^3 2x + log2^(x^2)`

= `(3)/(2)log(x^2 + 3) + 3log(sin2x) + x^2.log2`
Differentiating both sides w.r.t. x, we get
`(1)/y."dy"/"dx" = (3)/(2)"d"/"dx"[log(x^2 + 3)] + 3"d"/"dx"[log(sin2x)] + log2."d"/"dx"(x^2)`

= `(3)/(2) xx (1)/(x^2 + 3)."d"/"dx"(x^2 + 3) + 3 xx (1)/(sin2x)."d"/"dx"(sin2x) + log2 xx 2x`

= `(3)/(2(x^2 + 3)).(2x + 0) + (3)/(sin2x) xx cos2x."d"/"dx"(2x) + 2xlog2`

= `(6x)/(2(x^2 + 3)) + 3cot2x xx 2 + 2xlog2`

∴ `"dy"/"dx" = y[(3x)/(x^2 + 3) + 6cot2x + 2xlog2]`

= `(x^2 + 3)^(3/2).sin^3 2x.2^(x^2)[(3x)/(x^2 + 3) + 6cot2x + 2xlog2]`.