f(x)=\ln \left(x^{5}+x\right)-\frac{2}{\pi \operatorname{sen}\left(\frac{\pi x}{2}\right)} Calcular f^{\prime \prime}(1)

Passo 1

O enunciado dá
f(x)=\ln\bigl(x^{5}+x\bigr)\;-\;\frac{2}{\pi\,\sin\!\bigl(\tfrac{\pi x}{2}\bigr)}
e pede f''(1).
Vamos usar:
– derivada de \ln(u) pela regra da cadeia;
– derivada de 1/\sin(u) reconhecendo cosseno, cotangente e cosecante;
– regra do quociente para a primeira fração;
– depois avaliamos em x=1.

Passo 2

Calculamos f'(x).
Para a primeira parte, se
u(x)=x^{5}+x,
então
\frac{d}{dx}\ln\bigl(u(x)\bigr)=\frac{u'(x)}{u(x)} =\frac{5x^{4}+1}{x^{5}+x}\,.
A segunda parte é
-\,\frac{2}{\pi}\,\csc\!\bigl(\tfrac{\pi x}{2}\bigr).
Como \tfrac{d}{dx}\csc(u)=-\csc(u)\cot(u)\,u', com u=\tfrac{\pi x}{2} e u'=\tfrac{\pi}{2}, temos
\frac{d}{dx}\Bigl[-\tfrac{2}{\pi}\,\csc(u)\Bigr] =-\tfrac{2}{\pi}\Bigl[-\csc(u)\cot(u)\,\tfrac{\pi}{2}\Bigr] =\csc\!\bigl(\tfrac{\pi x}{2}\bigr)\,\cot\!\bigl(\tfrac{\pi x}{2}\bigr).
Logo,
f'(x) =\frac{5x^{4}+1}{x^{5}+x} \;+\;\csc\!\bigl(\tfrac{\pi x}{2}\bigr)\,\cot\!\bigl(\tfrac{\pi x}{2}\bigr).

Passo 3

Agora f''(x) é a soma das derivadas de cada parcela.
  1. Para
    \displaystyle A(x)=\frac{5x^{4}+1}{x^{5}+x}, use regra do quociente com
    \!A'(x)=\frac{(20x^{3})(x^{5}+x)-(5x^{4}+1)(5x^{4}+1)}{(x^{5}+x)^{2}}.
  2. Para
    \displaystyle B(x)=\csc(u)\cot(u) com u=\tfrac{\pi x}{2}, escrevemos
B'(x)=-u'\,\csc(u)\bigl[\cot^{2}(u)+\csc^{2}(u)\bigr] =-\tfrac{\pi}{2}\,\csc\!\bigl(\tfrac{\pi x}{2}\bigr)\Bigl[\cot^{2}\!\bigl(\tfrac{\pi x}{2}\bigr)+\csc^{2}\!\bigl(\tfrac{\pi x}{2}\bigr)\Bigr].
Portanto
f''(x) =\frac{20x^{3}(x^{5}+x)-(5x^{4}+1)^{2}}{(x^{5}+x)^{2}} \;-\;\frac{\pi}{2}\,\csc\!\bigl(\tfrac{\pi x}{2}\bigr)\Bigl[\cot^{2}\!\bigl(\tfrac{\pi x}{2}\bigr)+\csc^{2}\!\bigl(\tfrac{\pi x}{2}\bigr)\Bigr].

Passo 4

Avaliaremos em x=1. Note que \sin(\tfrac{\pi}{2})=1, \csc(\tfrac{\pi}{2})=1, \cot(\tfrac{\pi}{2})=0.
  1. Na parte racional:
20\cdot1^{3}(1^{5}+1)-(5\cdot1^{4}+1)^{2} =20\cdot2-6^{2}=40-36=4, \quad (x^{5}+x)^{2}\big|_{x=1}=(1+1)^{2}=4,
logo esse termo vale \tfrac{4}{4}=1.
2. Na parte trigonométrica:
-\frac{\pi}{2}\cdot1\bigl[0^{2}+1^{2}\bigr] =-\frac{\pi}{2}.
Somando, obtemos
f''(1)=1-\frac{\pi}{2}.

Resposta

f''(1)=1-\frac{\pi}{2}.

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