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phi
 one year ago
Best ResponseYou've already chosen the best response.2what is the derivative of \[ 7 x^{1} \] ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2using the power rule d/dx x^n = n x^(n1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wish I could tell you  the content doesn't cover that. It only gives me f(x) and ask me to find it.

phi
 one year ago
Best ResponseYou've already chosen the best response.2The rule is posted above. But here it is again \[ \frac{d}{dx} x^n = n x^{n1} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would variable "d" be then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The \(d\) is notation for an operator. Just like \(+\), it doesn't have a value

phi
 one year ago
Best ResponseYou've already chosen the best response.2d/dx is how you say "derivative with respect to x"

phi
 one year ago
Best ResponseYou've already chosen the best response.2the power rule for finding a derivative is one of the first things you learn in differential calculus. If this is mysterious, you need a refresher, because it is too much material to review here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm in Precalc, so we're probably never gonna cover the power rule. This is simply using the difference quotient.

phi
 one year ago
Best ResponseYou've already chosen the best response.2you mean using \[ f'(x) = \lim_{h\rightarrow0}\frac{ f(x+h)  f(x)}{(x+h)x} \]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, we're doing that to solve for the posted equation.

phi
 one year ago
Best ResponseYou've already chosen the best response.2in that case write \[ f'(x) = \lim_{h\rightarrow0}\frac{\frac{7}{x+h}  \frac{7}{x} }{h} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, sorry. Internet kinda cut out. I know we gotta find a common denominator between the two fractions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(which would be x for the first fraction, (x+h) for the second if I'm correct in saying that)

phi
 one year ago
Best ResponseYou've already chosen the best response.2or \[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h}  \frac{1}{x} \right)\]

phi
 one year ago
Best ResponseYou've already chosen the best response.2multiply the first fraction by x/x and the second by (x+h)/(x+h)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x }{ x^2 +xh}  \frac{ x+h }{ x^2+xh }\] Right?

phi
 one year ago
Best ResponseYou've already chosen the best response.2ok. now combine the top, and put the result over the common denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ h }{ x^2+xh }\] Alrightie.

phi
 one year ago
Best ResponseYou've already chosen the best response.2the top is x(x+h) = xxh

phi
 one year ago
Best ResponseYou've already chosen the best response.2\[ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h}  \frac{1}{x} \right) \\ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{h}{x^2+xh} \right) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2notice h/h divides out and you have \[ f'(x) = 7\lim_{h\rightarrow0} \left(\frac{1}{x^2+xh} \right) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2now let h "go to zero". the expression approaches 1/(x^2) you get \[ f'(x) = 7 \cdot \frac{1}{x^2} \] evaluate that at x=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So from here I'd replace x^2 with 1?

phi
 one year ago
Best ResponseYou've already chosen the best response.2you replace x with 1 so you get 1^2 or 1*1 in the denominator.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which gives the end result of 7?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, thank you so much for your help.
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