Question

Hello! High school student in Calculus here. I need help solving the problem attached in the image. I'd like help FINDING THE DERIVATIVE using PRODUCT RULE.I would greatly appreciate it, thank you! :)))

Answer 1

`(du)/(dm)=(\sec \text{ m)(}\csc m)\text{\lbrack(-cotm) +( tan }m)\rbrack`Step-by-step explanation:
`u(m)\text{ = secm}* co\sec m`

Using product rule:

dy/dx = V du/dx + u dv/dx

Here it becomes:

`(du)/(dm)=V(d(\csc m))/(dm)+U(d(\sec m))/(dm)`

V = sec m, u = csc m

`\begin{gathered} V\text{ = sec m} \\ \frac{dv}{dm\text{ }}=\sec m\text{ tanm} \end{gathered}`
`\begin{gathered} u\text{ = csc m} \\ (du)/(dm)=-\csc m\text{ cot m} \end{gathered}`

Inserting into the formula:

`\begin{gathered} (du)/(dm)=(\sec \text{ m)(-cscm cotm) +(csc m)(secm tan m)} \\ \end{gathered}`
`\begin{gathered} (du)/(dm)=(\sec \text{ m)\lbrack(-cscm) (cotm) +(csc m)( tan }m)\rbrack \\ (du)/(dm)=(\sec \text{ m)(}\csc m)\text{\lbrack(-cotm) +( tan }m)\rbrack \\ \end{gathered}`