# Math Help - differentiate and simplify #1

1. ## differentiate and simplify #1

$y={(7x-2)^{0.5}/(5x+3)}^{0.5}$

can someone check my work? thanks

$y'=0.5{(7x-2)/(5x+3)}^{-0.5}{(7)(5x+3)-(7x-2)(5)}/{(5x+3)^2}$
$y'=0.5{(7x-2)/(5x+3)}^{-0.5}{(31)/{(5x+3)^2}}$
$y'=(31)/{2(5x+3)^{1.5}(7x-2)^{0.5}}$

2. Originally Posted by skeske1234
$y={(7x-2)/(5x+3)}^{0.5}$

can someone check my work? thanks

$y'=0.5{(7x-2)/(5x+3)}^(-0.5){(7)(5x+3)-(7x-2)(5)}/{(5x+3)^2}$
$y'=0.5{(7x-2)/(5x+3)}^(-0.5){(31)/{(5x+3)^2}}$
$y'=(31)/{2(5x+3)^{1.5}(7x-2)^{0.5}}$

$y = \frac{7x-2}{\sqrt{5x+3}}
$

using the quotient rule ...

$y' = \frac{\sqrt{5x+3} \cdot 7 - (7x-2) \cdot \frac{5}{2\sqrt{5x+3}}}{5x+3}$

multiply numerator and denominator by $2\sqrt{5x+3}$ ...

$y' = \frac{14(5x+3) - 5(7x-2)}{2(5x+3)^{\frac{3}{2}}}$

$y' = \frac{35x + 52}{2(5x+3)^{\frac{3}{2}}}$

3. Originally Posted by skeeter
$y = \frac{7x-2}{\sqrt{5x+3}}
$

using the quotient rule ...

$y' = \frac{\sqrt{5x+3} \cdot 7 - (7x-2) \cdot \frac{5}{2\sqrt{5x+3}}}{5x+3}$

multiply numerator and denominator by $2\sqrt{5x+3}$ ...

$y' = \frac{14(5x+3) - 5(7x-2)}{2(5x+3)^{\frac{3}{2}}}$

$y' = \frac{35x + 52}{2(5x+3)^{\frac{3}{2}}}$

oops.. so sorry! the question was actually
$y={(7x-2)^{0.5}/(5x+3)}^{0.5}$

4. I'm not sure what method you have used.

I would apply the quoient rule which says $y= \frac{u}{v} \Rightarrow y' =\frac{vu'-uv'}{v^2}$

In your case $u = (7x-2)$ and $v = (5x+3)^{0.5}$

Now find $u'$ and $v'$ and you should be fine.

5. Originally Posted by skeske1234
oops.. so sorry! the question was actually
$y={(7x-2)^{0.5}/(5x+3)}^{0.5}$
... and your working is fine, wobbly latex aside

Edit: wow, pardon me for mis-approving the original post - lots of editing was going on... well, no - no excuse.

6. Originally Posted by skeske1234
oops.. so sorry! the question was actually
$y={(7x-2)^{0.5}/(5x+3)}^{0.5}$
you're not making this easy.

$y = \left(\frac{7x-2}{5x+3}\right)^{\frac{1}{2}}$

$y' = \frac{1}{2}\left(\frac{7x-2}{5x+3}\right)^{-\frac{1}{2}} \cdot \left(\frac{(5x+3)(7) - (7x-2)(5)}{(5x+3)^2}\right)
$

$y' = \frac{1}{2}\left(\frac{5x+3}{7x-2}\right)^{\frac{1}{2}} \cdot \left(\frac{31}{(5x+3)^2}\right)
$

$y' = \frac{31}{2(7x-2)^{\frac{1}{2}}(5x+3)^{\frac{3}{2}}}
$

7. Just in case a different (and I think comparably simple) route is of interest...

No quotient rule as such; we tweak a few balloons (on the bottom row) ...

... and we have the final denominator ready - just simplify the numerator.

_______________________
Don't integrate - balloontegrate!
Balloon Calculus Forum