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

$\frac{du}{dm}=(\sec \text{ m)(}\csc m)\text{\lbrack(-cotm) +( tan }m)\rbrack$Explanation:$u(m)\text{ = secm}\times co\sec m$

Using product rule:

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

Here it becomes:

$\frac{du}{dm}=V\frac{d(\csc m)}{dm}+U\frac{d(\sec m)}{dm}$

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} \\ \frac{du}{dm}=-\csc m\text{ cot m} \end{gathered}$

Inserting into the formula:

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