Question

cosx-senx/cosx?
qual a derivada básica?

Answer 1

Olá


$\displaystyle \mathsf{ \frac{cosx-senx}{cosx} }\\\\\\\text{Temos que aplicar a regra do quociente que diz o seguinte: }\\\\\mathsf{( \frac{f}{g} )'= \frac{f'\cdot g~- f\cdot g'}{g^2} }\\\\\\\\\mathsf{y'= \frac{(-senx- cosx)\cdot cosx~-~((cosx-senx)\cdot (-senx))}{(cosx)^2} }\\\\\\\text{Fazendo a distributiva}\\\\\\\mathsf{y'= \frac{(-senx\cdot cosx -cos^2x)~-~(-cosx\cdot senx+sen^2x)}{cos^2x} }\\\\\\\text{Distribui o sinal}$

$\displaystyle \mathsf{y'= \frac{-senx\cdot cosx -cos^2x+cosx\cdot senx -sen^2x}{cos^2x} }\\\\\\\text{Simplifica}\\\\\\\mathsf{y'= \frac{-~~\diagup\!\!\!\!\!\!\!\! senx\cdot~ \diagup\!\!\!\!\!\!\!cosx -cos^2x+~\diagup\!\!\!\!\!\!\!cosx\cdot~ \diagup\!\!\!\!\!\!\!senx -sen^2x}{cos^2x} }\\\\\\\mathsf{y'= \frac{-cos^2x-sen^2x}{cos^2x} }\\\\\\\text{Uma identidade trigonometrica basica diz que}\\\\\mathsf{cos^2x+sen^2x=1}\\\\\text{Se multiplicarmos a expressao por (-1) ficamos com}\\\\\mathsf{-cos^2x-sen^2x=-1}$

$\displaystyle \text{Vamos entao fazer essa substituicao}\\\\\mathsf{y'= \frac{-1}{cos^2x} }\\\\\\\text{Que podemos reescrever como }-sec^2x\\\\\\\boxed{\mathsf{y'=-sec^2x}}$