by HallsofIvy » Fri Jan 15, 2021 12:16 pm
First your function is ambiguous. What you wrote is [tex]sin(x)+ \frac{3}{cos(x)}[/tex] but I suspect you mean [tex]\frac{sin(x)+ 3}{cos(x)}[/tex].
Second, where did get this problem? I assume from a Calculus class but then you should know
1) that the derivative of sin(x) is cos(x)
2) that the derivative of cos(x) is -sin(x)
3) the quotient rule: [tex]\left(\frac{f}{g}\right)'= \frac{f'g- fg'}{g^2}[/tex].
If you mean [tex]\frac{sin(x)+ 3}{cos(x)}[/tex] then [tex]f(x)= sin(x)+ 3[/tex] and [tex]g(x)= cos(x)[/tex].
If really do you mean [tex]sin(x)+ \frac{3}{cos(x)}[/tex] then apply the quotient rule to [tex]\frac{3}{cos(x)}[/tex] with [tex]f(x)= 3[/tex] and [tex]g(x)= cos(x)[/tex].