Question

Differentiate (2tanx/tanx+cosx) ^2

Answer 1

we have to differentiate $\left(\frac{2tanx}{tanx+cosx}\right)^2$

let y = $\left(\frac{2tanx}{tanx+cosx}\right)^2$

now differentiating with respect to x,

dy/dx = 2[(2tanx)/(tanx + cosx)]²¯¹ d{2tanx/(tanx + cosx)}/dx

= 4tanx/(tanx + cosx) [(tanx + cosx) d(2tanx)/dx - 2tanx d(tanx + cosx)/dx ]/(tanx + cosx)²

= 4tanx/(tanx + cosx)³ [(tanx + cosx)(2sec²x) - 2tanx(sec²x - sinx)]

= 4tanx/(tanx + cosx)³ [2tanx. sec²x + 2secx - 2tanx. sec²x + 2tanx . sinx]

= 4tanx/(tanx + cosx)³ [2secx + 2tanx. sinx ]

= 4tanx/(tanx + cosx)³ [2secx + 2secx. sin²x ]

= 4tanx/(tanx + cosx)³ [2secx (1 + sin²x)]

= {4tanx × [2secx (1 + sin²x)]} /(tanx + cosx)³

= 8tanx . secx(1 + sin²x)/(tanx + cosx)³

therefore differentiation of $\left(\frac{2tanx}{tanx+cosx}\right)^2$is $\frac{8tanx secx(1+sin^2x)}{(tanx+cosx)^3}$