unimatix
  • unimatix
Derivatives without using shortcut method. Problem: g(x) = x sqrt(x)
Mathematics
jamiebookeater
  • jamiebookeater
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unimatix
  • unimatix
Going to show what I have as of now: \[g(x) = xsqrt(x)\] \[y = xsqrt(x)\] \[y + \triangle y = x + \triangle(x)\sqrt{ (x+\triangle x ) )}\]
unimatix
  • unimatix
|dw:1441578438052:dw|
triciaal
  • triciaal
|dw:1441579069139:dw|

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triciaal
  • triciaal
|dw:1441579234601:dw|
triciaal
  • triciaal
this is going too far back
unimatix
  • unimatix
I know the product rule. I'm trying to do it without though.
unimatix
  • unimatix
I tried to rationalize the numerator on the left and got nothing.
Jhannybean
  • Jhannybean
\[g(x) = x\sqrt{x}\]\[g(x+h) = (x+h)(\sqrt{x+h})\]\[\frac{g(x+h)-g(x)}{h}=\frac{(x+h)(\sqrt{x+h})-x\sqrt{x}}{h} \]
Jhannybean
  • Jhannybean
Meh.. the only thing that'll work here would be : \[=\frac{\sqrt{(x+h)^3}-x^{3/2}}{h}= \frac{(x+h)^{3/2}-x^{3/2}}{h} \]
amistre64
  • amistre64
you are trying to work first principles .. the limit definition of a derivative right?
Jhannybean
  • Jhannybean
Mmhmm... \(\lim_{h\rightarrow 0}\)
amistre64
  • amistre64
(x+h) sqrt(x+h) - x sqrt(x) ------------------------- h x sqrt(x+h) +hsqrt(x+h) - x sqrt(x) ------------------------------- h x (sqrt(x+h) - sqrt(x)) sqrt(x+h) + ------------------- h x (h) sqrt(x+h) + ------------------- h(sqrt(x+h) + sqrt(x)) x sqrt(x+h) + ------------------- sqrt(x+h) + sqrt(x))
amistre64
  • amistre64
do you follow it?
Jhannybean
  • Jhannybean
Not the third step.
amistre64
  • amistre64
(ac+b)/a = c + (b/a)
amistre64
  • amistre64
you just split the fraction into convenient terms, and the left side is just a simplified result of h/h sqrt(x+h)
Jhannybean
  • Jhannybean
Oh, I see what you did now.
amistre64
  • amistre64
\[\frac{a+bh+c}{h}\color{red}{\implies}\frac{bh}{h}+\frac{a+c}{h}\]
amistre64
  • amistre64
the rest is taken care of by a conjugate
Jhannybean
  • Jhannybean
Just trying to solve it by myself real quick and cross-checking between my work and your method :)
Jhannybean
  • Jhannybean
Then.... \[\lim_{h\rightarrow 0} \left(\sqrt{x+h} +\frac{x}{2\sqrt{x}}\right) \]\[=\sqrt{x}+\frac{x}{2\sqrt{x}}\]\[=\frac{2x+x}{2\sqrt{x}}\]\[=\frac{3x}{2\sqrt{x}}\]\[=\frac{3x^{3/2}}{2x}\]\[=\boxed{\frac{3}{2}x^{1/2}}\]

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