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# Derivation help!

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$$f'(x)=(x-2)/\sqrt{x}$$

How do I get the answer x+2/2xsqrtx ?

Dec 1, 2015

#4
+10

Ok  Namodesto

I have made some changes to the original post, let me know how you get on.

Why did I divide by x:

I used the quotient rule.  The denominator of the quotient rule is v^2

v=sqrt{x}

v^2 = x

Anyway let me know if youdo or if you don't understand :))

Dec 2, 2015

#1
+15

The question is not written properly Namodesto

It should be like this

$$f(x)=\frac{x+2}{\sqrt{x}}\\ \mbox{Use quotient rule}\\ \boxed{If\;\;y=\frac{u(x)}{v(x)}\;\;then\;\;y'=\frac{vu'-uv'}{v^2} }\\ u=x+2 \quad v=x^{0.5}\\ u'=1 \qquad v'=0.5v^{-0.5}=\frac{1}{2\sqrt2}\\$$

$$(1)\qquad f'(x)=\frac{\sqrt{x}-\frac{1}{2\sqrt{x}}(x-2)}{x}\\ (2)\qquad f'(x)=\left[\sqrt{x}-\frac{1}{2\sqrt{x}}(x-2)\right] \div x\\ (3) \qquad f'(x)=\left[\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (4) \qquad f'(x)=\left[\frac{2\sqrt{x}}{2\sqrt{x}}*\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (5) \qquad f'(x)=\left[\frac{2\sqrt{x}}{2\sqrt{x}}*\frac{\sqrt{x}}{1}-\frac{(x-2)}{2\sqrt{x}}\right] \div x\\ (6) \qquad f'(x)=\left[\frac{2\sqrt{x}\sqrt{x}}{2\sqrt{x}*1}-\frac{(x-2)}{2\sqrt{x}}\right] \times \frac{1}{x}\\(7) \qquad f'(x)=\left[\frac{2x}{2\sqrt{x}}-\frac{(x-2)}{2\sqrt{x}}\right] \times \frac{1}{x}\\ (8) \qquad f'(x)=\frac{2x-(x-2)}{2\sqrt{x}}\times \frac{1}{x}\\ (9) \qquad f'(x)=\frac{2x-(x-2)}{2x\sqrt{x}}\\ (10) \qquad f'(x)=\frac{x+2}{2x\sqrt{x}}\\$$

I have added lines to help you understand and I have put line numbers so you can tell me which bit you do not understand.

So let me know how you get on :)

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Dec 2, 2015
edited by Melody  Dec 2, 2015
edited by Melody  Dec 2, 2015
#2
0

Hi Melody,

thank you so much for your help, however I still don't understand how you got 2x. And also why do you divide by x?

Dec 2, 2015
#4
+10

Ok  Namodesto

I have made some changes to the original post, let me know how you get on.

Why did I divide by x:

I used the quotient rule.  The denominator of the quotient rule is v^2

v=sqrt{x}

v^2 = x

Anyway let me know if youdo or if you don't understand :))

Melody Dec 2, 2015
#5
+5

Thank you Melody! I finally understand haha :)

Dec 2, 2015
#6
+5

That is great.  I am glad I could help :)

Dec 2, 2015