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Section 3.5 : Derivatives of Trig Functions
10. Differentiate \(\displaystyle Z\left( v \right) = \frac{{v + \tan \left( v \right)}}{{1 + \csc \left( v \right)}}\) .
Show SolutionNot much to do here other than take the derivative, which will require the quotient rule.
\[\begin{align*}Z'\left( v \right) & = \frac{{\left( {1 + {{\sec }^2}\left( v \right)} \right)\left( {1 + \csc \left( v \right)} \right) - \left( {v + \tan \left( v \right)} \right)\left( { - \csc \left( v \right)\cot \left( v \right)} \right)}}{{{{\left( {1 + \csc \left( v \right)} \right)}^2}}}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\left( {1 + {{\sec }^2}\left( v \right)} \right)\left( {1 + \csc \left( v \right)} \right) + \csc \left( v \right)\cot \left( v \right)\left( {v + \tan \left( v \right)} \right)}}{{{{\left( {1 + \csc \left( v \right)} \right)}^2}}}}}\end{align*}\]