I am trying to find the derivative of (x-5)sqr(x+2)

[qazwsx]

I have posted the solution but have no clue were the 2 comes from.

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\(\displaystyle \begin{align} D_x y\, &=\, (x\, -\, 5)\, D_x\, \sqrt{\strut x\, +\, 2\,}\, +\, \left(\sqrt{\strut x\, +\, 2\,}\right)\, D_x\, (x\, -\, 5)

\\ \\ &=\, (x\, -\, 5)\, \left(\dfrac{1}{2}\right)\, (x\, +\, 2)^{-\frac{1}{2}}\, (1)\, +\, (x\, +\, 2)^{\frac{1}{2}}\, (1)

\\ \\ &=\, \dfrac{1}{2}\, (x\, -\, 5)\, (x\, +\, 2)^{-\frac{1}{2}}\, +\, (x\, +\, 2)^{\frac{1}{2}}

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, \bigg[(x\, -\, 5)\, +\, \color{red}{\large{2}}\, (x\, +\, 2)\bigg]

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, (3x\, -\, 1) \end{align}\)

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I [highlighted] the 2 that is in question

Thankls

 

[stapel]

I have posted the solution but have no clue were the 2 comes from.

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\(\displaystyle \begin{align} D_x y\, &=\, (x\, -\, 5)\, D_x\, \sqrt{\strut x\, +\, 2\,}\, +\, \left(\sqrt{\strut x\, +\, 2\,}\right)\, D_x\, (x\, -\, 5)

\\ \\ &=\, (x\, -\, 5)\, \left(\dfrac{1}{2}\right)\, (x\, +\, 2)^{-\frac{1}{2}}\, (1)\, +\, (x\, +\, 2)^{\frac{1}{2}}\, (1)

\\ \\ &=\, \dfrac{1}{2}\, (x\, -\, 5)\, (x\, +\, 2)^{-\frac{1}{2}}\, +\, (x\, +\, 2)^{\frac{1}{2}}

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, \bigg[(x\, -\, 5)\, +\, \color{red}{\large{2}}\, (x\, +\, 2)\bigg]

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, (3x\, -\, 1) \end{align}\)

Well, what did you get when you factored the negative-1/2-power radical out of both terms? You started with the basic algebra:

. . . . .\(\displaystyle \dfrac{1}{2}\, (x\, -\, 5)\, (x\, +\, 2)^{-\frac{1}{2}}\, +\, (x\, +\, 2)^{\frac{1}{2}}\, =\, \dfrac{1}{2}\, (x\, -\, 5)\, \dfrac{1}{\sqrt{\strut x\, +\, 2\,}}\, +\, \dfrac{x\, +\, 2}{\sqrt{\strut x\, +\, 2\,}}\, =\, ...\)

And... then what?

Please reply showing all of your steps so far. Thank you!

 

[qazwsx]

why is (x+2)^1/2 in the denominator in your exmple. If that was the case I could multiply by the conjugate but that doesnt seem correct to me.

 

[Deleted member 4993]

I have posted the solution but have no clue were the 2 comes from.

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\(\displaystyle \begin{align} D_x y\, &=\, (x\, -\, 5)\, D_x\, \sqrt{\strut x\, +\, 2\,}\, +\, \left(\sqrt{\strut x\, +\, 2\,}\right)\, D_x\, (x\, -\, 5)

\\ \\ &=\, (x\, -\, 5)\, \left(\dfrac{1}{2}\right)\, (x\, +\, 2)^{-\frac{1}{2}}\, (1)\, +\, (x\, +\, 2)^{\frac{1}{2}}\, (1)

\\ \\ &=\, \dfrac{1}{2}\, (x\, -\, 5)\, (x\, +\, 2)^{-\frac{1}{2}}\, +\, \color{red}{\large{1}}\,(x\, +\, 2)^{\frac{1}{2}}

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, \bigg[(x\, -\, 5)\, +\, \color{red}{\large{2}}\, (x\, +\, 2)\bigg]

\\ \\ &=\, \dfrac{1}{2}\, (x\, +\, 2)^{-\frac{1}{2}}\, (3x\, -\, 1) \end{align}\)

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I [highlighted] the 2 that is in question

Thankls

The line before - you have a silent '1'. When you try to factor out ½ from that you get 2.

 

[qazwsx]

The line before - you have a silent '1'. When you try to factor out ½ from that you get 2.

ok the silent one make complete sense. I was overthinking this way to much. Thank you for the help