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Did I do the following derivatives correctly?

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

I used the quotient rule and got

$\frac{(x-5)(12x^3+2)-(1)(3x^4+2x)}{(x-5)^2}$ apparently I do not have to simplify

my second question

find the derivative of

$f(x)=(2x^5-3x^2)(\sqrt{x}-4x)$

I used product rule and got

$f(x)'=(2x^5-3x^2)(\frac{1}{2\sqrt{x}}-4)+(10x^4-6x)(\sqrt{x}-4x)$

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If looks good, except you forgot the $-4$ in the derivative of $\sqrt{x}-4x$. –  Daryl Feb 2 '13 at 21:38
    
I did not notice that –  Fernando Martinez Feb 2 '13 at 21:40

1 Answer 1

up vote 2 down vote accepted
  • Your first derivative looks fine.

Find the derivative of $f(x)=(2x^5-3x^2)(\sqrt{x}-4x)$

"I used product rule and got: $f(x)'=(2x^5-3x^2)(\frac{1}{2\sqrt{x}})+(10x^4-6x)(\sqrt{x}-4x)$"

  • You did fine; the process of deriving is correct. The only mis-step is you omitted a term of $-4$ when deriving $\sqrt x - 4x$

So, corrected, you should have: $$f(x)'=(2x^5-3x^2)\left(\frac{1}{2\sqrt{x}}-4\right)+(10x^4-6x)(\sqrt{x}-4x)$$

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I have a quick question would not the $-4$ be separate from $\fract{1}\2sqrt{x} because to get it on the numerator do you have to multiply. –  Fernando Martinez Feb 2 '13 at 21:51
    
thanks for the compliment.You are sharp too. –  Fernando Martinez Feb 2 '13 at 21:52

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