Help (Four to Seven) Thank you

How to find the derivative of this function?

Lelechka [254]

$\bf y=\cfrac{x^3-2x+1}{\sqrt{x}}\implies y=\cfrac{x^3-2x+1}{x^{\frac{1}{2}}}
\\\\\\
\cfrac{dy}{dx}=\stackrel{quotient~rule}{\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x})^2}}
\\\\\\
\cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~(x^3-2x+1)\cdot\frac{1}{2\sqrt{x}}}{x}$

$\bf \cfrac{dy}{dx}=\cfrac{(3x^2-2)\sqrt{x}~~~-~~~\cdot\frac{x^3-2x+1}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{[(3x^2-2)\sqrt{x}](2\sqrt{x})~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{\frac{[(3x^2-2)2x]~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}
\implies 
\cfrac{dy}{dx}=\cfrac{\frac{[6x^3-4x]~~-~~(x^3-2x+1)}{2\sqrt{x}}}{x}$

$\bf \cfrac{dy}{dx}=\cfrac{\frac{6x^3-4x~~-~~x^3+2x-1}{2\sqrt{x}}}{x}\implies 
\cfrac{dy}{dx}=\cfrac{\frac{5x^3-2x-1}{2\sqrt{x}}}{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x\sqrt{x}}$

$\bf \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^1\cdot x^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{1+\frac{1}{2}}}\\\\\\ \cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2x^{\frac{3}{2}}}
\implies
\cfrac{dy}{dx}=\cfrac{5x^3-2x-1}{2\sqrt{x^3}}$

8 0

3 years ago

To compare and order fractions,the fraction can be written with a

Greeley [361]

Common denominator i believe. i’m not sure. hope this helps

5 0

2 years ago

The company financial officer was interested in the average cost of PCs that had been purchased in the past six months. A random

enot [183]

Answer:

2995

Step-by-step explanation:

The first quartile (Q1) is defined as the middle number between the smallest number and the median of the data set

1127

1482

2995

3009

3250

3445

3449

4000

6120

rearranging the data sets from the least to the highest

the middle number of the data sets is 3250

the middle number between the smallest number and the median of the data set is 2995

7 0

3 years ago

What is the difference in length between 1 1/4 inch button and a 3/8 inch button

jeka94

You have two buttons of different lengths.

First with length $1\dfrac{1}{4}$ in and second with length $\dfrac{3}{8}$ in. The first one is larger, then the difference in length between them is: