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MIT 18.01 Single Variable Calculus (OCW) 16 Online
OpenStudy (anonymous):

how do you find the derivative of f(x)=xsinx+1/cosx?

OpenStudy (noelgreco):

Can you put some parentheses on the right-hand side? It's ambiguous.

OpenStudy (anonymous):

NoelGreco is right, you gotta clarify that

OpenStudy (anonymous):

If you mean (xsinx+1)/cosx,then use the rule of divide to get ((cosx)(xcosx+sinx)-(xsinx+1)(-sinx))/(cosx)^2 If you mean xsinx+ 1/cosx,then,find the derivative seperately.Namely,(xcosx+sinx)+secxtanx

OpenStudy (anonymous):

product rule for the first part and quotient for the second

OpenStudy (anonymous):

\[\frac{ x }{ \cos ^{2}(x) } + tanx (\frac{ \cos x -1}{ \cos x } )\]

OpenStudy (anonymous):

sinx+cosx-(sinx/cos^2x)

OpenStudy (anonymous):

hmmmm... first use product rule for xsinx... then 1/cosx = secx.... find derivative of secx.. job done

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