Chapter 3.4, Problem 67E

To determine

To find: The 50th derivative of $y=\cos 2x$.

Answer to Problem 67E

The 50th derivative of $y=\cos 2x$ is $\underset{\_}{\frac{d^{50}y}{d{x}^{50}}=-2^{50}\cos 2x}$.

Explanation of Solution

Given:

The function is $y=\cos 2x$.

Calculation:

The first derivative of $y=\cos 2x$ is computed as follows,

$\begin{array}{c}{y}^{\prime }=\frac{d}{dx}\left(y\right)\\ =\frac{d}{dx}\left(\cos 2x\right)\\ =-\sin 2x\cdot \left(2\right)\\ =-2\sin 2x\end{array}$

Thus, the first derivative of $y=\cos 2x$ is $y^{\prime }=-2\sin 2x$.

The second derivative of $y$ is computed as follows,

$\begin{array}{c}{y}^{″}=\frac{d}{dx}\left(y^{\prime }\right)\\ =\frac{d}{dx}\left(-2\sin 2x\right)\\ =-2\cdot \frac{d}{dx}\left(\sin 2x\right)\end{array}$

Simplify the terms,

$\begin{array}{c}{y}^{″}=-2\left(\frac{d}{d\left(2x\right)}\left(\sin 2x\right)\cdot \frac{d\left(2x\right)}{dx}\right)\\ =-2\left(\cos 2x\cdot 2\right)\\ =-2^2\left(\cos 2x\right)\end{array}$

Thus, the second derivative of $y=\cos 2x$ is $y^{″}=-2^2\left(\cos 2x\right)$.

The third derivative of y is computed as follows,

$\begin{array}{c}{y}^{‴}=\frac{d}{dx}\left(y^{″}\right)\\ =\frac{d}{dx}\left(-2^2\left(\cos 2x\right)\right)\\ =-2^2\frac{d}{dx}\left(\left(\cos 2x\right)\right)\\ =-2^2\left(-\sin 2x\cdot 2\right)\end{array}$

     $=2^3\left(\sin 2x\right)$

Thus, the third derivative of $y=\cos 2x$ is $y^{‴}=2^3\left(\sin 2x\right)$.

Proceed in the similar way, the 50th derivative y is computed as follows,

Since 50 = 12(4) + 2, the remainder is 2 and it corresponds with $y^{″}$.

$\frac{d^{50}y}{d{x}^{50}}=-2^{50}\cos 2x$

Therefore, the 50th derivative of $y=\cos 2x$ is $\underset{\_}{\frac{d^{50}y}{d{x}^{50}}=-2^{50}\cos 2x}$.