Chapter 2.3, Problem 33E

To determine

To calculate: The derivative $\frac{dy}{dx}$ of the function

Answer to Problem 33E

The derivative of the function is. $y^{\prime }=4x^3+3x^2-4x+1$ , $y^{″}=12x^2+6x-4$ , $y^{‴}=24x+6$ , $y^4=24$

Explanation of Solution

Given information:

The function is $y=x^4+x^3-2x^2+x-5$ .

Concept used:

The quotient rule is $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^2}$ and the formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

The function is $y=x^4+x^3-2x^2+x-5$ .

Differentiate the function with respect to $x$ for the first order derivative.

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^4+x^3-2x^2+x-5\right)\\ =\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(x^3\right)-2\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(5\right)\\ =4x^3+3x^2-2\times 2x+1+0\\ =4x^3+3x^2-4x+1\end{array}$

Differentiate the function with respect to $x$ for the second order derivative.

  $\begin{array}{c}\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(4x^3+3x^2-4x+1\right)\\ =4\frac{d}{dx}\left(x^3\right)+3\frac{d}{dx}\left(x^2\right)-4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)\\ =4.\left(3x^2\right)+3.\left(2x\right)-4\left(1\right)+0\\ =12x^2+6x-4\end{array}$

Differentiate the function with respect to $x$ for the third order derivative.

  $\begin{array}{c}\frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=\frac{d}{dx}\left(12x^2+6x-4\right)\\ =12\frac{d}{dx}\left(x^2\right)+6\frac{d}{dx}\left(x\right)-4\frac{d}{dx}\left(1\right)\\ =12.\left(2x\right)+6.\left(1\right)-4\left(0\right)\\ =24x+6\end{array}$

Differentiate the function with respect to $x$ for the fourth order derivative.

  $\begin{array}{c}\frac{d}{dx}\left(\frac{d^{3}y}{d{x}^4}\right)=\frac{d}{dx}\left(24x+6\right)\\ =24\frac{d}{dx}\left(x\right)+6\frac{d}{dx}\left(1\right)\\ =24+0\\ =24\end{array}$

Conclusion: The derivative of the function is. $y^{\prime }=4x^3+3x^2-4x+1$ , $y^{″}=12x^2+6x-4$ , $y^{‴}=24x+6$ , $y^4=24$ .