Derive as funções abaixo:
f(x)=\frac{\operatorname{arctg}(5 x+1)}{x} ; g(x)=\frac{x^{2} \operatorname{arcsen}(x)}{x+1} ; h(x)=\operatorname{arcsen}(\operatorname{arctg}(2 x)) .

Passo 1

O enunciado pede para calcular as derivadas de
f(x)=\frac{\operatorname{arctg}(5x+1)}{x},\quad g(x)=\frac{x^{2}\,\operatorname{arcsen}(x)}{x+1},\quad h(x)=\operatorname{arcsen}\bigl(\operatorname{arctg}(2x)\bigr).
Para isso vamos usar:
  1. a derivada de \operatorname{arctg}(u) é \dfrac{u'}{1+u^{2}};
  2. a derivada de \operatorname{arcsen}(u) é \dfrac{u'}{\sqrt{1-u^{2}}};
  3. regra do quociente e regra do produto;
  4. regra da cadeia.

Passo 2

Derivando f(x)=\dfrac{\operatorname{arctg}(5x+1)}{x} pela regra do quociente:
se u(x)=\operatorname{arctg}(5x+1) e v(x)=x, então
u'(x)=\frac{5}{1+(5x+1)^{2}},\quad v'(x)=1
e
f'(x) =\frac{v(x)\,u'(x)-u(x)\,v'(x)}{[v(x)]^{2}} =\frac{x\cdot\frac{5}{1+(5x+1)^{2}}-\operatorname{arctg}(5x+1)}{x^{2}}.

Passo 3

Derivando
g(x)=\frac{x^{2}\,\operatorname{arcsen}(x)}{x+1}
pela regra do quociente: escreva a(x)=x^{2}\,\operatorname{arcsen}(x) e b(x)=x+1.
Então
a'(x)=2x\,\operatorname{arcsen}(x)+x^{2}\cdot\frac{1}{\sqrt{1-x^{2}}},\quad b'(x)=1,
e
g'(x) =\frac{a'(x)\,b(x)-a(x)\,b'(x)}{[b(x)]^{2}} =\frac{\bigl(2x\,\operatorname{arcsen}(x)+\tfrac{x^{2}}{\sqrt{1-x^{2}}}\bigr)(x+1) -x^{2}\,\operatorname{arcsen}(x)}{(x+1)^{2}}.

Passo 4

Derivando
h(x)=\operatorname{arcsen}\bigl(\operatorname{arctg}(2x)\bigr)
pela regra da cadeia em duas etapas:
\displaystyle u(x)=\operatorname{arctg}(2x) com
u'(x)=\frac{2}{1+(2x)^{2}}=\frac{2}{1+4x^{2}},
e depois \operatorname{arcsen}(u)
h'(x) =\frac{u'(x)}{\sqrt{1-[u(x)]^{2}}} =\frac{\frac{2}{1+4x^{2}}}{\sqrt{1-\bigl[\operatorname{arctg}(2x)\bigr]^{2}}} =\frac{2}{(1+4x^{2})\,\sqrt{1-\bigl[\operatorname{arctg}(2x)\bigr]^{2}}}.

Resposta

f'(x) =\frac{\displaystyle x\;\frac{5}{1+(5x+1)^{2}}-\operatorname{arctg}(5x+1)}{x^{2}},
g'(x) =\frac{\bigl(2x\,\operatorname{arcsen}(x)+\tfrac{x^{2}}{\sqrt{1-x^{2}}}\bigr)(x+1) -x^{2}\,\operatorname{arcsen}(x)}{(x+1)^{2}},
h'(x) =\frac{2}{(1+4x^{2})\,\sqrt{1-\bigl[\operatorname{arctg}(2x)\bigr]^{2}}}.

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