Question

Find the 50^th derivative of \(y = \cos 2x\).

Short answer

The 50^th derivative of the function is \({D^{50}}\cos 2x = - {2^{50}}\cos 2x\).

Step-by-step solution

The Chain Rule

The composite function \(F = f \circ g\), or \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiableat \(x\) and \(F'\) is obtained as:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

In Leibniz notation, the derivative is shown below:

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Determine the 50th derivative of the function

The inner function is \(u = g\left( x \right) = 2x\) , and the outer function is \(y = f\left( u \right) = \cos u\).

Use the notation \(D,{D^2}, \ldots ,{D^n}\) to represent the derivative. In particular, the derivative is \(Df\left( {2x} \right) = 2f'\left( {2x} \right),{D^2}f\left( {2x} \right) = 4f''\left( {2x} \right), \ldots ,\)\({D^n}f\left( {2x} \right) = {2^n}{f^{\left( n \right)}}\left( {2x} \right)\).

As the function \(f\left( x \right) = \cos x\) and \(50 = 4\left( {12} \right) + 2\) we obtain \({f^{\left( {50} \right)}}\left( x \right) = {f^{\left( 2 \right)}}\left( x \right) = - \cos x\). Therefore, \({D^{50}}\cos 2x = - {2^{50}}\cos 2x\).

Thus, the 50^th derivative of the function is \({D^{50}}\cos 2x = - {2^{50}}\cos 2x\).