Differentiate (2tanx/tanx+cosx) ^2

1.	Differentiate (2tanx/tanx+cosx) ^2
Answer» we have to DIFFERENTIATE $\left(\frac{2tanx}{tanx+<klux>COSX</klux>}\right)^2$ 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 $\left(\frac{2tanx}{tanx+cosx}\right)^2$ therefore DIFFERENTIATION of is $\frac{8tanx secx(1+sin^2x)}{(tanx+cosx)^3}$

Answer»

we have to DIFFERENTIATE $\left(\frac{2tanx}{tanx+<klux>COSX</klux>}\right)^2$

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

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

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