Derivar e Siuplificar
y=log e 1+tg x21-tg x2

Passo 1

O enunciado pede a derivada de y = lnl(tfrac1+tan(tfracx2)1-tan(tfracx2)r). Vamos usar regra da cadeia para logaritmo natural e derivada de tangente, além da regra do quociente para funções trigonométricas.

Passo 2

Definimos u = tfrac1+tan(tfracx2)1-tan(tfracx2) e calculamos u' pela regra do quociente. Seja
N = 1 + tanl(tfracx2r), D = 1 - tanl(tfracx2r),
as derivadas são
$$N’ = \tfrac{1}{2}\sec^2\bigl(\tfrac{x}{2}\bigr),\quad D’ = -\tfrac{1}{2}\sec^2\bigl(\tfrac{x}{2}\bigr).
$$

Passo 3

Então
$$u’ = \frac{N’D - N D’}{D^2} = \frac{\tfrac12\sec^2(\tfrac{x}{2})(1-\tan(\tfrac{x}{2})) - (1+\tan(\tfrac{x}{2}))\bigl(-\tfrac12\sec2(\tfrac{x}{2})\bigr)}{(1-\tan(\tfrac{x}{2}))2} = \frac{\sec2(\tfrac{x}{2})}{(1-\tan(\tfrac{x}{2}))2}.
$$

Passo 4

Agora derivamos y=ln(u) usando y'=u'/u:
$$y’ = \frac{u’}{u} = \frac{\tfrac{\sec2(\tfrac{x}{2})}{(1-\tan(\tfrac{x}{2}))2}}{\tfrac{1+\tan(\tfrac{x}{2})}{1-\tan(\tfrac{x}{2})}} = \frac{\sec^2(\tfrac{x}{2})}{(1-\tan(\tfrac{x}{2}))(1+\tan(\tfrac{x}{2}))} = \frac{\sec2(\tfrac{x}{2})}{1-\tan2(\tfrac{x}{2})}.
$$

Passo 5

Para simplificar usamos identidades trigonométricas:
1-tan^2l(tfracx2r) = cos^2l(tfracx2r) - sin^2l(tfracx2r) = cos(x),
e também sec^2(tfracx2)=1/cos^2(tfracx2). Logo
$$y’ = \frac{1/\cos2(\tfrac{x}{2})}{\cos(x)/\cos2(\tfrac{x}{2})} = \frac{1}{\cos(x)} = \sec(x).
$$

Resposta

y’ = sec(x)

Exercícios relacionados