Question

Se y = (ln x)^cos x determine dy/dx

Answer 1

podemos usar o bizu de fazer o seguinte :

$\displaystyle \text y = (\text{ln x})^{\text{cos(x)}}$

vamos aplicar ln dos dois lados :

$\text{ln}( \text y) = \text{ln(ln x)}^{\text{cos(x)}} \\\\ \text{ln(y) = \text{cos(x).\text{ln(ln x) }}}$

Agora vamos derivar :

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

$\displaystyle \text{y '}= \text{y}. ( -\text{sen(x)}.\text{ln(ln x) }+\frac{\text{cos(x)}}{\text{(ln x).x}}\ )$

Agora é só substituir o valor de y  :

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