Question

If y = `("e"^"2x" sin x)/(x cos x), "then" "dy"/"dx" = ?`

Options

• `("e"^"2x"[(2x - 1) tan x - x sec^2 x])/x^2`

• `("e"^"2x"[(2x + 1) tan x - x sec^2 x])/x^2`

• `("e"^"2x"[(2x - 1) tan x + x sec^2 x])/x^2`

• None of these

Answer 1

`("e"^"2x"[(2x - 1) tan x + x sec^2 x])/x^2`

Explanation:

y = `("e"^"2x" sin x)/(x cos x)`

Taking logarithm on both sides, we get

log y = 2x + log (sin x) - log x - log (cos x)

Differentiating w.r.t. x, we get

`1/"y" * "dy"/"dx" = 2 + ((cos x)/(sin x)) - 1/x + (sin x)/(cos x)`

`=> "dy"/"dx" = ("e"^"2x" sin x)/(x cos x) (2 + (cos x)/(sin x) - 1/x + (sin x)/(cos x))`

`= "e"^"2x" (2/x tan x + 1/x - 1/x^2 tan x + (tan^2x)/x)`

`= "e"^"2x"/x^2 [2x tan x - tan x + x(1 + tna^2 x)]`

`= "e"^"2x"/x^2 [(2x - 1) tan x + x sec^2 x]`