Chapter 4, Problem 43RE

To determine

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

Answer to Problem 43RE

The derivative is $\frac{d^{40}y}{d{x}^{40}}=32e^{x\sqrt[8]{2}}$ .

Explanation of Solution

Given Information: The given equation is $y=e^{x\sqrt[8]{2}}$ .

Calculation:

The derivative of the function is,

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

Differentiate further,

  $\begin{array}{c}\frac{dy}{dx}=\sqrt[8]{2}{e}^{x\sqrt[8]{2}}\\ \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\sqrt[8]{2}{e}^{x\sqrt[8]{2}}\right)\\ \frac{d^{2}y}{d{x}^2}=\sqrt[8]{2}\frac{d}{dx}\left(e^{x\sqrt[8]{2}}\right)\left(\text{Take the constany out}\right)\\ \frac{d^{2}y}{d{x}^2}=\sqrt[8]{2}\times \sqrt[8]{2}{e}^{x\sqrt[8]{2}}\left(\because \frac{d}{dx}\left(e^{nx}\right)=n{e}^{nx}\right)\\ \frac{d^{2}y}{d{x}^2}={\left(\sqrt[8]{2}\right)}^2{e}^{x\sqrt[8]{2}}\end{array}$

Differentiate further,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}={\left(\sqrt[8]{2}\right)}^2{e}^{x\sqrt[8]{2}}\\ \frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=\frac{d}{dx}\left({\left(\sqrt[8]{2}\right)}^2{e}^{x\sqrt[8]{2}}\right)\\ \frac{d^{3}y}{d{x}^3}={\left(\sqrt[8]{2}\right)}^2\frac{d}{dx}\left(e^{x\sqrt[8]{2}}\right)\left(\text{Take the constany out}\right)\\ \frac{d^{3}y}{d{x}^3}={\left(\sqrt[8]{2}\right)}^2\times \sqrt[8]{2}{e}^{x\sqrt[8]{2}}\left(\because \frac{d}{dx}\left(e^{nx}\right)=n{e}^{nx}\right)\\ \frac{d^{3}y}{d{x}^3}={\left(\sqrt[8]{2}\right)}^3{e}^{x\sqrt[8]{2}}\end{array}$

From above derivatives, a pattern is observed. The nth derivative of $y=e^{x\sqrt[8]{2}}$ is n -times $\left(\sqrt[8]{2}\right)$ multiplied by $e^{x\sqrt[8]{2}}$ .

Hence, the 40^th derivative of $y=e^{x\sqrt[8]{2}}$ is,

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

The derivative is $\frac{d^{40}y}{d{x}^{40}}=32e^{x\sqrt[8]{2}}$ .