Chapter 5.5, Problem 17E

To determine

To calculate:

The value of $dy$ and evaluate its value in $y=x^2\ln x$

Answer to Problem 17E

The value of $dy$ is $\left(2x\ln x+x\right)dx$ and $0.01$ at the given values

Explanation of Solution

Given information:

  $x=1$

  $dx=0.01$

Formula used:

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

  $\begin{array}{l}y=x^2\ln x\\ dy=\left(2x\ln x+x\right)dx\\ dy=\left\{2\left(1\right)\ln \left(1\right)+1\right\}\left(0.01\right)\\ dy=\left(2\times 0+1\right)\left(0.01\right)\\ dy=1\times 0.01\\ dy=0.01\end{array}$

Therefore, the value of $dy$ is $\left(2x\ln x+x\right)dx$ and $0.01$ at the given values