MHB Need help with these derivatives

[rainbowbaconz]

http://www.mathway.com/math_image.aspx?p=f%28x%29SMB01%28SMB02RSMB03xSMB02rSMB03+SMB02FSMB031SMB10SMB02RSMB03xSMB02rSMB03SMB02fSMB03%29SMB02ESMB033SMB02eSMB03?p=117?p=42, show that http://www.mathway.com/math_image.aspx?p=SMB02OSMB03fSMB04,xSMB02oSMB03SMB01SMB02FSMB033xSMB02ESMB033SMB02eSMB03+3xSMB02ESMB032SMB02eSMB03-3x-3SMB102xSMB02ESMB032SMB02eSMB03SMB02RSMB03xSMB02rSMB03SMB02fSMB03?p=165?p=42

http://www.mathway.com/math_image.aspx?p=f%28x%29SMB01xSMB02ESMB032SMB02eSMB03%28xSMB02ESMB032SMB02eSMB03-1%29%28xSMB02ESMB033SMB02eSMB03-1%29?p=168?p=22, show that http://www.mathway.com/math_image.aspx?p=SMB02OSMB03fSMB04,xSMB02oSMB03SMB01x%28x-1%29%287xSMB02ESMB034SMB02eSMB03+7xSMB02ESMB033SMB02eSMB03+2xSMB02ESMB032SMB02eSMB03-2x-2%29?p=273?p=22

http://www.mathway.com/math_image.aspx?p=f%28x%29SMB01SMB02RSMB03%28SMB02FSMB031-SMB02RSMB03xSMB02rSMB03SMB101+SMB02RSMB03xSMB02rSMB03SMB02fSMB03%29SMB02rSMB03?p=113?p=42, show that http://www.mathway.com/math_image.aspx?p=SMB02OSMB03fSMB04,xSMB02oSMB03SMB01SMB02FSMB033xSMB02ESMB033SMB02eSMB03+3xSMB02ESMB032SMB02eSMB03-3x-3SMB102xSMB02ESMB032SMB02eSMB03SMB02RSMB03xSMB02rSMB03SMB02fSMB03?p=165?p=42

http://www.mathway.com/math_image.aspx?p=f%28x%29SMB01SMB02FSMB032xSMB02ESMB03SMB02FSMB033SMB102SMB02fSMB03SMB02eSMB03-xSMB02ESMB03SMB02FSMB035SMB102SMB02fSMB03SMB02eSMB03+3SMB02RSMB03xSMB02rSMB03SMB10-SMB02RSMB03xSMB02rSMB03SMB02fSMB03?p=137?p=42, show that http://www.mathway.com/math_image.aspx?p=SMB02OSMB03fSMB04,xSMB02oSMB03SMB012%28x-1%29?p=107?p=22

http://www.mathway.com/math_image.aspx?p=f%28x%29SMB01SMB02FSMB03SMB02RSMB031+xSMB02rSMB03SMB10SMB02RSMB031-xSMB02rSMB03SMB02fSMB03?p=88?p=42, show that http://www.mathway.com/math_image.aspx?p=SMB02OSMB03fSMB04,xSMB02oSMB03SMB01SMB02FSMB031SMB10%281-x%29SMB02RSMB031-xSMB02ESMB032SMB02eSMB03SMB02rSMB03SMB02fSMB03?p=139?p=42
I know how to get the first derivatives for all these problems, but I'm having problem getting to these alternate derivative forms. Please help. Thanks.

 

[Jameson]

Hi rainbowbaconz,

Welcome to MHB :)

We usually ask for one problem per thread, so which one shall we start with? Pick one and then show us what you did and we'll help you keep going.

Jameson

 

[MarkFL]

Hello and welcome, rainbowbaconz!

It is to your advantage to show what your thoughts are and/or the work you have so far so that we may specifically address where you may be going astray. This helps you in all future problems of this type, since you will be able "fix" the error in the application of the rules for differentiation.

We want to help you to understand the problems, and then to be able to apply the rules to other problems with success.

And please don't feel afraid to be wrong, no one here will think any less of you. We have all been there, trust me! The only thing to be embarrassed about is to be unsure and not ask for guidance.

 

[soroban]

Hello, rainbowbaconz!

The fourth one is elementary . . .

f(x) \:=\:\dfrac{2x^{\frac{3}{2}} - x^{\frac{5}{2}} + 3\sqrt{x}}{\text{-}\sqrt{x}}

\text{Show that: }\:f'(x) \:=\:2(x-1)

$\text{We have: }\:f(x) \;=\;\dfrac{2x^{^{\frac{3}{2}}} - x^{^{\frac{5}{2}}} + 3x^{^{\frac{1}{2}}}}{\text{-}x^{^{\frac{1}{2}}}} $

. . . . . . . $ f(x)\;=\;\dfrac{2x^{\frac{3}{2}}}{\text{-}x^{\frac{1}{2}}} - \dfrac{x^{\frac{5}{2}}}{\text{-}x^{\frac{1}{2}}} + \dfrac{3x^{\frac{1}{2}}}{\text{-}x^{\frac{1}{2}}} $

. . . . . . . $f(x) \;=\;\text{-}2x + x^2 - 3$Hence: ..$f'(x) \;=\;\text{-}2 + 2x \;=\;2(x-1)$

 

[alane1994]

Here, refer to this.

http://www.mathhelpboards.com/misc.php?do=vsarules

These are some of the expectations when you make a thread or post.