Question

me ajudem nesses 2 itens, por favor.​

Answer 1

$h(\theta)=cossec(\theta)+e^{\theta}.cotg(\theta)\\\\h'(\theta)=[cossec(\theta)]'+[e^{\theta}]'.cotg(\theta)+e^{\theta}.[cotg(\theta)]'\\\\h'(\theta)=-cossec(\theta).cotg(\theta)+e^{\theta}.cotg(\theta)+e^{\theta}(-cossec^2(\theta))\\\\h'(\theta)=-cossec(\theta).cotg(\theta)+e^{\theta}(cotg(\theta)-cossec^2(\theta))$

$f(\theta)=\frac{sec(\theta)}{1+sec(\theta)}\\\\f'(\theta)=\frac{[sec(\theta)]'[1+sec(\theta)]-[sec(\theta)][1+sec(\theta)]'}{[1+sec(\theta)]^2}\\\\f'(\theta)=\frac{[sec(\theta).tg(\theta)][1+sec(\theta)]-sec(\theta)[sec(\theta).tg(\theta)]}{[1+sec(\theta)]^2}\\\\f'(\theta)=\frac{[sec(\theta).tg(\theta)][1+sec(\theta)-sec(\theta)]}{[1+sec(\theta)]^2}\\\\f'(\theta)=\frac{sec(\theta).tg(\theta)}{[1+sec(\theta)]^2}$

Se desejar simplificar:

$f'(\theta)=\frac{\frac{1}{cos(\theta)}\frac{sen(\theta)}{cos(\theta)}}{[1+\frac{1}{cos(\theta)}]^2}\\\\f'(\theta)=\frac{\frac{sen(\theta)}{cos^2(\theta)}}{\frac{[cos(\theta)+1]^2}{[cos(\theta)]^2}}\\\\f'(\theta)=\frac{sen(\theta)}{[1+cos(\theta)]^2}$