- #1
Chemical_Penguin
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Hi all, I've been trying to get back into mathematics by teaching myself calculus. I've been starting with the book "Calculus Made Easy" and have been doing fine except for one little thing I encountered on page 57.
He shows a mathematical proof of why the derivative of
[tex]y = x^\frac {1} {2} [/tex]
is...
[tex] \frac {dy} {dx} = \frac {1} {2}x^\frac {-1} {2}[/tex]
But there's one step that I don't understand and would greatly appreciate if someone could explain it for me. Here it is:
[tex] y + dy = \sqrt {x} (1 + \frac {dx} {x})^\frac {1} {2} [/tex]
To...
[tex]= \sqrt {x} + \frac {1} {2} \frac {dx} {\sqrt {x}} - \frac {1} {8} \frac {dx^2} {x \sqrt {x}} +... [/tex]"terms with higher powers of dx"
The last part "terms with higher powers of dx" is exactly how this problem is listed in the book. If anyone at all could explain to me what exactly is going on between these steps I would be so happy!
-Thanks in advance.
-CP
He shows a mathematical proof of why the derivative of
[tex]y = x^\frac {1} {2} [/tex]
is...
[tex] \frac {dy} {dx} = \frac {1} {2}x^\frac {-1} {2}[/tex]
But there's one step that I don't understand and would greatly appreciate if someone could explain it for me. Here it is:
[tex] y + dy = \sqrt {x} (1 + \frac {dx} {x})^\frac {1} {2} [/tex]
To...
[tex]= \sqrt {x} + \frac {1} {2} \frac {dx} {\sqrt {x}} - \frac {1} {8} \frac {dx^2} {x \sqrt {x}} +... [/tex]"terms with higher powers of dx"
The last part "terms with higher powers of dx" is exactly how this problem is listed in the book. If anyone at all could explain to me what exactly is going on between these steps I would be so happy!
-Thanks in advance.
-CP
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