Question

7-52 Find the derivative of the function.

\(y = {\bf{sin}}\left( {\theta + {\bf{tan}}\left( {\theta + {\bf{cos}}\theta } \right)} \right)\)

Short answer

The derivative of the given equation is \(\cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)} \right) \times \left( {1 - \sin \theta } \right)\).

Step-by-step solution

Recall the chain rule

According to the chain rule, the derivative of a function \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) can be obtained as:

\(\begin{array}{c}F'\left( x \right) = f'\left( {g\left( x \right)} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {g\left( x \right)} \right)\\ = f'\left( {g\left( x \right)} \right)g'\left( x \right)\end{array}\)

Find the derivative of the given function 

The derivative of the equation \(y = \sin \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}\theta }} &= \frac{{\rm{d}}}{{{\rm{d}}\theta }}\left( {\sin \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)} \right)\\ &= \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right) \times \frac{{\rm{d}}}{{{\rm{d}}\theta }}\left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\\ &= \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)} \right) \times \frac{{\rm{d}}}{{{\rm{d}}\theta }}\left( {\theta + \cos \theta } \right)\\ &= \cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)} \right) \times \left( {1 - \sin \theta } \right)\end{aligned}\)

Thus, the derivative of the given equation is \(\cos \left( {\theta + \tan \left( {\theta + \cos \theta } \right)} \right)\left( {1 + {{\sec }^2}\left( {\theta + \cos \theta } \right)} \right) \times \left( {1 - \sin \theta } \right)\).