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Derivative of a fraction

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  • #1
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Homework Statement


if $$y = \frac{2x^5-3x^3+x^2}{x^3}$$ then $$\frac{dy}{dx} =$$

Homework Equations


if $$f(x) = x^n$$ then $$f'(x) = nx^{n-1}$$

The Attempt at a Solution


$$\frac{2x^5-3x^3+x^2}{x^3} = \frac{2x^5}{x^3} - \frac{3x^3}{x^3} + \frac{x^2}{x^3}$$
$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}$$
$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - 3 + \frac{2x}{3x^2}$$

And now I am stuck as to how to simplify this. Should I have done the whole polynomial division before I took the derivative?
 

Answers and Replies

  • #2
SteamKing
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Homework Statement


if $$y = \frac{2x^5-3x^3+x^2}{x^3}$$ then $$\frac{dy}{dx} =$$

Homework Equations


if $$f(x) = x^n$$ then $$f'(x) = nx^{n-1}$$

The Attempt at a Solution


$$\frac{2x^5-3x^3+x^2}{x^3} = \frac{2x^5}{x^3} - \frac{3x^3}{x^3} + \frac{x^2}{x^3}$$
$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}$$
$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - 3 + \frac{2x}{3x^2}$$

And now I am stuck as to how to simplify this. Should I have done the whole polynomial division before I took the derivative?
The derivative of [f(x) / g (x)] ≠ f'(x) / g'(x).

If you have the expression 2x5/x3, how would you simplify that before taking the derivative?
 
  • #3
Doc Al
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$$\frac{2x^5-3x^3+x^2}{x^3} = \frac{2x^5}{x^3} - \frac{3x^3}{x^3} + \frac{x^2}{x^3}$$
This is fine.

$$ f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}$$
This is not!

Look up the quotient rule.

And now I am stuck as to how to simplify this. Should I have done the whole polynomial division before I took the derivative?
Yes.
 
  • #4
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I read a site with derivative shortcuts and it said this "if you have a polynomial with only 1 term in the denominator than you can separate it into individual fractions and take the derivative of each of those fractions separately.

If I were to divide each term by x3 than I would get 2x2 - 3 + 1/x
 
  • #5
Doc Al
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I read a site with derivative shortcuts and it said this "if you have a polynomial with only 1 term in the denominator than you can separate it into individual fractions and take the derivative of each of those fractions separately.

If I were to divide each term by x3 than I would get 2x2 - 3 + 1/x
Nothing wrong with that. Now you can take the derivative of each term separately.
 
  • #6
SteamKing
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Look up the quotient rule.
I think the OP is just starting to learn derivatives and he only knows the power rule.

I think the purpose of this exercise was to show how the original expression could be split up and simplified before applying the power rule. Unfortunately, the OP did not simplify and thought (incorrectly) that the derivative of a quotient was the quotient of the derivatives.
 
  • #7
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Also not mentioned by the other folks responding in this thread is the incorrect use of notation.
##f'(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}##
This does NOT mean the derivative of the quantity in parentheses. It means to evaluate f' at ##\frac{2x^5-3x^3+x^2}{x^3}##. A better use of notation would be the following:
##\frac{d}{dx}(\frac{2x^5-3x^3+x^2}{x^3}) = \frac{d}{dx}(\frac{10x^4}{3x^2} - \frac{9x^2}{3x^2} + \frac{2x}{3x^2}) = ... ##
 
  • #8
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But how is d/dx any different from f'? They are just 2 notations for the same thing, the first derivative. Same with 2 of these d/dx and f'' for second derivative and so on for theoretically infinitely many derivatives.
 
  • #9
SteamKing
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But how is d/dx any different from f'? They are just 2 notations for the same thing, the first derivative. Same with 2 of these d/dx and f'' for second derivative and so on for theoretically infinitely many derivatives.
Typically, the derivative is expressed as follows: for a function y = f(x), y' = f'(x) = df(x)/dx, where x is the independent variable. x does not represent, for example, another function, which is the distinction that Mark44 was trying to make.

If f(x) = x2, then writing f(x2) should mean (x2)2 to be consistent with functional notation.
 
  • #10
Doc Al
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Unfortunately, the OP did not simplify and thought (incorrectly) that the derivative of a quotient was the quotient of the derivatives.
I agree, of course. (Which is why I suggested that he look up the Quotient Rule for derivatives, so he knows why what he did was wrong.)
 
  • #11
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But how is d/dx any different from f'? They are just 2 notations for the same thing, the first derivative. Same with 2 of these d/dx and f'' for second derivative and so on for theoretically infinitely many derivatives.
No, they are not the same. In addition to what SteamKing said, the operator d/dx indicates that the goal is to take the derivative of what follows it. The function f' is the derivative of some function f. It does not mean to take the derivative of what is in the parentheses.
 

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