Question

7-52 Find the derivative of the function.

\(y = {\bf{cos}}\sqrt {{\bf{sin}}\left( {{\bf{tan}}\,\pi x} \right)} \)

Short answer

The derivative of the given equation is \(\frac{{ - \pi \sin \sqrt {\sin \left( {\tan \pi x} \right)} \left( {\cos \left( {\tan \pi x} \right)} \right){{\sec }^2}\pi x}}{{2\sqrt {\sin \left( {\tan \pi x} \right)} }}\).

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:

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

Find the derivative of the given function

The derivative of the equation \(y = \cos \sqrt {\sin \left( {\tan \pi x} \right)} \).

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

Thus, the derivative of the given equation is\(\frac{{ - \pi \sin \sqrt {\sin \left( {\tan \pi x} \right)} \left( {\cos \left( {\tan \pi x} \right)} \right){{\sec }^2}\pi x}}{{2\sqrt {\sin \left( {\tan \pi x} \right)} }}\).