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Hey, I just need help with this problem. I'm not sure about the working out, can someone please help me out?

2x5-3x4+6x3-2x2 /3x2

 

The answer is 2x2-2x+2

Thank you 

Ashreeta  Aug 14, 2017
 #1
avatar+93691 
+1

2x5-3x4+6x3-2x2 /3x2

 

\(\quad\frac{d}{dx}\;\;\frac{2x^5-3x^4+6x^3-2x^2 }{3x^2}\\ =\frac{d}{dx}\;\;\left[\frac{2x^5}{3x^2} - \frac{3x^4}{3x^2} + \frac{ 6x^3}{3x^2} - \frac{2x^2 }{3x^2}\right]\\ =\frac{d}{dx}\;\;\left[\frac{2x^3}{3} - x^2 + 2x - \frac{2 }{3}\right]\\ =\frac{3*2x^2}{3} - 2x + 2 - 0\\ =2x^2-2x+2\)

Melody  Aug 14, 2017
 #2
avatar+20035 
+1

Hey, I just need help with this problem. I'm not sure about the working out, can someone please help me out?

Differentiate\( \frac{2x^5-3x^4+6x^3-2x^2 }{3x^2}\)

The answer is \(2x^2-2x+2\)

Thank you 

 

\(\begin{array}{|rcll|} \hline y &=& \frac{2x^5-3x^4+6x^3-2x^2 }{3x^2} \\ &=& \frac{x^2 \cdot \left( 2x^3-3x^2+6x-2\right) }{3x^2} \\ &=& \frac{ 2x^3-3x^2+6x-2 }{3} \\ &=& \frac23 x^3-x^2+2x-\frac23 \\\\ y' & = & \frac23\cdot 3x^2 -2x + 2 \\ \mathbf{y'} & \mathbf{=} & \mathbf{2x^2 -2x + 2} \\ \hline \end{array}\)

 

laugh

heureka  Aug 14, 2017
edited by heureka  Aug 14, 2017
 #3
avatar+70 
+1

Thank you Melody and Heureka 

:) 

It helped a lot 

Ashreeta  Aug 14, 2017

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