Question

Calcule a derivada de cada uma das funções abaixo.

Answer 1

item a)

$\displaystyle \text y=\text{cos}(\text x^2-1)\text{ln}(\sqrt{\text x}) \\\\ \underline{\text{Derivando}}: \\\\ \text y'=[\text{cos}(\text x^2-1)]'.\text{ln}(\sqrt{\text x})+\text{cos}(\text x^2-1).[\text{ln}(\sqrt{\text x})]' \\\\ \text y'= -\text{sen}(\text x^2-1).(\text x^2-1)'.\text{ln}(\sqrt{\text x})+\text{cos}(\text x^2-1).\frac{(\sqrt{\text x})'}{\sqrt{\text x}}$

$\displaystyle \text y'=-\text{sen}(\text x^2-1)(2\text x).\text{ln}(\sqrt{\text x})+\text{cos}(\text x^2-1)\frac{1}{2.\sqrt{\text x}.\sqrt{\text x}} \\\\\\\ \boxed{\text y'=-2\text x.\text{sen}(\text x^2-1)\text{ln}(\sqrt{\text x})+\frac{\text{cos}(\text x^2-1)}{2\text x}\ }$

item b)

$\displaystyle \text y = \text{arcsen}(\text x^2+\text x) \\\\ \underline{\text{Vamos relembra a derivada do arcsen(u) (onde u {\'e} uma fun{\c c}{\~a}o)}}: \\\\ \text{y}'=[\text{arcsen(u)}]' \\\\ \text y'= \frac{\text u'}{\sqrt{1-\text u^2 }} \\\\\\ \underline{\text{temos}}: \\\\ \text y'=[\text{arcseno}(\text x^2+\text x)]' \\\\ \text y' = \frac{(\text x^2+\text x)'}{\sqrt{1-(\text x^2+\text x)^2}} \\\\\\\ \huge\boxed{\text y'=\frac{2\text x+1}{\sqrt{1-(\text x^2+\text x)^2}} \ }\checkmark$

item c)

$\displaystyle \text y = (\text x+1)^{\displaystyle \text{sen (x)}} \\\\\ \underline{\text{Vamos aplicar ln dos dois lados para derivar melhor}}: \\\\ \text{ln y}=\text{ln}(\text x+1)^{\displaystyle \text{sen (x)}} \\\\ \text{ln y}=\text{sen(x)}.\text{ln(x+1)} \\\\ \underline{\text{Derivando}}: \\\\ \frac{\text y'}{\text y} = [\text{sen(x)}]'.\text{ln(x+1)}+\text{sen(x)}[\text{ln(x+1)}]' \\\\\\ \frac{\text y'}{\text y} = \text{cos(x)}.\text{ln(x+1)}+\frac{\text{sen(x)}}{\text{x+1}}$

$\displaystyle \text y'=\text y[ \ \text{cos(x)}.\text{ln(x+1)}+\frac{\text{sen(x)}}{\text x+1} \ ] \\\\\\ \boxed{\text y'=(\text x+1)^{\displaystyle \text{sen(x)}} [\ \text{cos(x)}.\text{ln(x+1)}+\frac{\text{sen(x)}}{\text x+1} \ ] \ }\checkmark$

item d)

$\displaystyle \text{y}=\frac{\text{tan(x)}}{\text{cos(3x)}} \\\\\\ \text y'=\frac{[\text{tan(x)}]'.\text{cos(3x)+\text{tan(x)}[\text{cos(3x)}]'}}{\ [\text{cos(3x)}\ ]^2} \\\\\\ \text y'=\frac{\text{sec}^2(\text x).\text{cos(3x)}+\text{tan(x)}[-\text{sen(3x)}3]}{\text{cos}^2(3\text x)} \\\\\\ \text y'=\frac{\text{sec}^2(\text x).\text{cos(3x)}-3.\text{tan(x).sen(3x)}}{\text{cos}^2(3\text x)}\ }$

$\displaystyle \boxed{\text y'=\text{sec}^2(\text x).\text{sec(3x)}-3\text{tan(x)}.\text{tan(3x)}.\text{sec(3x)} \ }\checkmark$