Chapter 4, Problem 44RE

To determine

To calculate: To find $\frac{d^{40}y}{d{x}^{40}}$ .a

Answer to Problem 44RE

The derivative is $32\sin \left(\sqrt[8]{2}x\right)$ .

Explanation of Solution

Given Information: The given equation is $y=\sin \left(\sqrt[8]{2}x\right)$ .

Calculation:

The derivative of the function is,

  $\begin{array}{c}y=\sin \left(\sqrt[8]{2}x\right)\\ \frac{dy}{dx}=\frac{d}{dx}\left(\sin \left(\sqrt[8]{2}x\right)\right)\left(\text{Substitute the value of }y\right)\\ \frac{dy}{dx}=\cos \left(\sqrt[8]{2}x\right)\frac{d}{dx}\left(\sqrt[8]{2}x\right)\left(\because \frac{d}{dx}\left(\sin x\right)=\cos x\right)\\ \frac{dy}{dx}=\sqrt[8]{2}\cos \left(\sqrt[8]{2}x\right)\end{array}$

Differentiate further,

  $\begin{array}{c}\frac{dy}{dx}=\sqrt[8]{2}\cos \left(\sqrt[8]{2}x\right)\\ \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\sqrt[8]{2}\cos \left(\sqrt[8]{2}x\right)\right)\left(\text{Substitute the value of }\frac{dy}{dx}\right)\\ \frac{d^{2}y}{d{x}^2}=\sqrt[8]{2}\frac{d}{dx}\left(\cos \left(\sqrt[8]{2}x\right)\right)\left(\text{Take the constant out}\right)\\ \frac{d^{2}y}{d{x}^2}=\sqrt[8]{2}\times -\sin \left(\sqrt[8]{2}x\right)\frac{d}{dx}\left(\sqrt[8]{2}x\right)\left(\because \frac{d}{dx}\left(\cos x\right)=-\sin x\right)\\ \frac{d^{2}y}{d{x}^2}=-{\left(\sqrt[8]{2}\right)}^2\sin \left(\sqrt[8]{2}x\right)\end{array}$

Differentiate further,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=-{\left(\sqrt[8]{2}\right)}^2\sin \left(\sqrt[8]{2}x\right)\\ \frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=\frac{d}{dx}\left(-{\left(\sqrt[8]{2}\right)}^2\sin \left(\sqrt[8]{2}x\right)\right)\\ \frac{d^{3}y}{d{x}^3}=-{\left(\sqrt[8]{2}\right)}^2\frac{d}{dx}\left(\sin \left(\sqrt[8]{2}x\right)\right)\left(\text{Take the constant out}\right)\\ \frac{d^{3}y}{d{x}^3}=-{\left(\sqrt[8]{2}\right)}^2\times \cos \left(\sqrt[8]{2}x\right)\frac{d}{dx}\left(\sqrt[8]{2}x\right)\left(\because \frac{d}{dx}\left(\sin x\right)=\cos x\right)\\ \frac{d^{3}y}{d{x}^3}=-{\left(\sqrt[8]{2}\right)}^3\cos \left(\sqrt[8]{2}x\right)\end{array}$

Differentiate further,

  $\begin{array}{c}\frac{d^{3}y}{d{x}^3}=-{\left(\sqrt[8]{2}\right)}^3\cos \left(\sqrt[8]{2}x\right)\\ \frac{d}{dx}\left(\frac{d^{3}y}{d{x}^3}\right)=\frac{d}{dx}\left(-{\left(\sqrt[8]{2}\right)}^3\cos \left(\sqrt[8]{2}x\right)\right)\\ \frac{d^{4}y}{d{x}^4}=-{\left(\sqrt[8]{2}\right)}^3\frac{d}{dx}\left(\cos \left(\sqrt[8]{2}x\right)\right)\left(\text{Take the constant out}\right)\\ \frac{d^{4}y}{d{x}^4}=-{\left(\sqrt[8]{2}\right)}^3\times -\sin \left(\sqrt[8]{2}x\right)\frac{d}{dx}\left(\sqrt[8]{2}x\right)\left(\because \frac{d}{dx}\left(\cos x\right)=-\sin x\right)\\ \frac{d^{4}y}{d{x}^4}={\left(\sqrt[8]{2}\right)}^4\sin \left(\sqrt[8]{2}x\right)\end{array}$

From above derivatives, a pattern is observed. After every 4^thderivative the sign of the derivative becomes positive and the nth derivative of $y=\sin \left(\sqrt[8]{2}x\right)$ is n -times $\left(\sqrt[8]{2}\right)$ multiplied by $\sin \left(\sqrt[8]{2}x\right)$ . It can also be observed that the mth -derivative of the function where m is even is takes sin value and when m is odd, it takes cosine value. Since 40 is divisible by 4, the sign of the 40^th derivative will be positive and the value is,

  $\begin{array}{c}\frac{d^{40}y}{d{x}^{40}}={\left(\sqrt[8]{2}\right)}^{40}\sin \left(\sqrt[8]{2}x\right)\\ ={\left(2^{\frac{1}{8}}\right)}^{40}\sin \left(\sqrt[8]{2}x\right)\\ =2^5\sin \left(\sqrt[8]{2}x\right)\\ \frac{d^{40}y}{d{x}^{40}}=32\sin \left(\sqrt[8]{2}x\right)\end{array}$

The derivative is $32\sin \left(\sqrt[8]{2}x\right)$ .