Derive as funções trigonométricas inversas a seguir:
f(x)=\operatorname{arcsen}\left(\ln \left(x^{5}+x^{2}\right)\right)+\operatorname{arccos} \left(e^{x^{2}} \cos \left(x^{2}-x\right)\right)

Passo 1

O enunciado pede a derivada de
f(x)=\arcsin\bigl(\ln(x^{5}+x^{2})\bigr)\;+\;\operatorname{arccos}\bigl(e^{x^{2}}\cos(x^{2}-x)\bigr).
Vamos aplicar a regra da cadeia e as fórmulas \frac{d}{dx}\bigl[\arcsin(u)\bigr]=\frac{u'}{\sqrt{1-u^{2}}}
e \frac{d}{dx}\bigl[\operatorname{arccos}(v)\bigr]=-\frac{v'}{\sqrt{1-v^{2}}}.

Passo 2

Defina
u(x)=\ln\bigl(x^{5}+x^{2}\bigr).
Então
\frac{d}{dx}\bigl[\arcsin(u(x))\bigr] =\frac{u'(x)}{\sqrt{1-[u(x)]^{2}}}.
Falta achar u'(x).

Passo 3

Usando \frac{d}{dx}[\ln(w)]=w'/w com w=x^{5}+x^{2}, temos
u'(x) =\frac{5x^{4}+2x}{\,x^{5}+x^{2}\,}.
Logo a derivada do primeiro termo é
\frac{5x^{4}+2x}{\bigl(x^{5}+x^{2}\bigr)\,\sqrt{1-\bigl[\ln(x^{5}+x^{2})\bigr]^{2}}}.

Passo 4

Agora defina
v(x)=e^{x^{2}}\cos(x^{2}-x).
Então
\frac{d}{dx}\bigl[\operatorname{arccos}(v(x))\bigr] =-\frac{v'(x)}{\sqrt{1-[v(x)]^{2}}}.
Falta achar v'(x).

Passo 5

Para v(x)=e^{x^{2}}\cdot\cos(x^{2}-x) usamos produto e cadeia:
v'(x) =2x\,e^{x^{2}}\cos(x^{2}-x) +e^{x^{2}}\bigl[-\sin(x^{2}-x)\bigr]\,(2x-1).
Simplificando,
v'(x) =e^{x^{2}}\bigl(2x\cos(x^{2}-x)-(2x-1)\sin(x^{2}-x)\bigr).
Então a derivada do segundo termo é
-\frac{e^{x^{2}}\bigl(2x\cos(x^{2}-x)-(2x-1)\sin(x^{2}-x)\bigr)} {\sqrt{1-\bigl[e^{x^{2}}\cos(x^{2}-x)\bigr]^{2}}}.

Passo 6

Somando as duas partes,
f'(x) =\frac{5x^{4}+2x}{\bigl(x^{5}+x^{2}\bigr)\,\sqrt{1-\bigl[\ln(x^{5}+x^{2})\bigr]^{2}}} \;-\;\frac{e^{x^{2}}\bigl(2x\cos(x^{2}-x)-(2x-1)\sin(x^{2}-x)\bigr)} {\sqrt{1-\bigl[e^{x^{2}}\cos(x^{2}-x)\bigr]^{2}}}.

Resposta

f'(x) =\frac{5x^{4}+2x}{\bigl(x^{5}+x^{2}\bigr)\,\sqrt{1-\bigl[\ln(x^{5}+x^{2})\bigr]^{2}}} \;-\;\frac{e^{x^{2}}\bigl(2x\cos(x^{2}-x)-(2x-1)\sin(x^{2}-x)\bigr)} {\sqrt{1-\bigl[e^{x^{2}}\cos(x^{2}-x)\bigr]^{2}}}.

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