MHB 1.3 find dx of (7+9x-6\sqrt{x})/x

[karush]

Find $\dfrac{dy}{dx}$ for:$y=\dfrac{7+9x-6\sqrt{x}}{x}$

ok several ways to solve this but thot the best might be to expand it first so

$y=\dfrac{7}{x} +9 -6\dfrac{\sqrt{x}}{x}$

or is there a better way

 

[topsquark]

Find $\dfrac{dy}{dx}$ for:$y=\dfrac{7+9x-6\sqrt{x}}{x}$

ok several ways to solve this but thot the best might be to expand it first so

$y=\dfrac{7}{x} +9 -6\dfrac{\sqrt{x}}{x}$

or is there a better way

It depends on what form you want for your answer. If you want to do it fast I'd recommend dividing through first. If you want it simplified (ie. as a fraction) it would be faster to use the quotient rule. But it's really a matter of taste.

-Dan

 

[karush]

$\left(\dfrac{f}{g}\right)'=\dfrac{f'\cdot g-g'\cdot f}{g^2}$
quotient rule looks kinda busy since
$f'=-\frac{3}{\sqrt{x}}+9$

but expansion wasn't easy either
$\left(\dfrac{7}{x}\right)'=-\dfrac{7}{x^2}$
$\quad 9'=0$
$\quad\left(-6\cdot \dfrac{\sqrt{x}}{x}\right)'
=\dfrac{3}{x^{\dfrac{3}{2}}}$
$\implies -\dfrac{7}{x^2}+\dfrac{3}{x^{\dfrac{3}{2}}}$

hence $y'=\dfrac{3\sqrt{x}-7}{x^2}$

hopefully

 

[topsquark]

hence $y'=\dfrac{3\sqrt{x}-7}{x^2}$

hopefully

You've been doing much more involved problems over the years. You don't need the "hopefully." If you doubt your answer, check it by integration!

Yes. that's the derivative.

-Dan

 

[Kansas Boy]

This is, I think, simpler- write the function as $y= 7x^{-1}+ 9- 6x^{-1/2}$. Now use the fact that the derivative of $x^n$ is $nx^{n- 1}$

$\frac{dy}{dx}= -7x^{-2}- 3x^{-3/2}$ which can be written in the original form as $\frac{dy}{dx}= \frac{7-3\sqrt{x}}{x^2}$​