Question

Differentiate the following w.r.t. x : `"cosec"^-1[(10)/(6sin(2^x) - 8cos(2^x))]`

Answer 1

Let y = `"cosec"^-1[(10)/(6sin(2^x) - 8cos(2^x))]`

= `sin^-1[(6sin(2^x) - 8cos(2^x))/(10)]    ...[∵ "cosec"^-1 x = sin^-1(1/x)]`

= `sin^-1[{sin(2^x)}(6/10) - {cos(2^x)}(8/10)]`

Since, `(6/10)^2 + (8/10)^2 = (36)/(100) +(64)/(100)` = 1,

we can write, `(6)/(10) = cos∞ and (8)/(10) = sin∞`.

∴ y = sin^–1[sin(2^x).cos∞ – cos(2^x).sin∞]
= sin^–1[sin(2^x – ∞)]
= 2^x – ∞, where ∞ is a constant
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(2^x - ∞ )`

= `"d"/"dx"(2^x) - "d"/"dx"(∞)`
= 2^x.log2 – 0
= 2^x.log2