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$\begingroup$

$y={\dfrac{[3x+1]}{[4x-1]^\frac{1}{2}}}$

Is it logical to square both the numerator and denominator to cancel out the power to $\frac12$ in the denominator?

Edit: This expression is being used for a chain rule application.

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    It's the square root only pertinent to the denominator? – 2017-01-19
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    Yes. That's the way the question is set up. – 2017-01-19
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    If you square one side, you need to square the other side. Are you trying to differentiate? – 2017-01-19
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    What exactly do you want to do? Just get rid of the square root? Then yes, of course you can square the expression. – 2017-01-19
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    This expression is for a chain rule application. Having a square root in the denominator while trying to calculate dy/dx doesn't make the process easy. Thus, would it be reasonable to square on both sides of the fraction to make calculating dy/dx simpler? – 2017-01-19
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    If you want to differentiate, I believe logarithmic differentiation is the best way to proceed. – 2017-01-19
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    Note that $\dfrac{a^2}{b^2} = \left(\dfrac ab\right )^2$ – 2017-01-19

2 Answers 2

2

Assuming those are meant to be normal parenthesis around the numerator and denominator, squaring both top and bottom of the fraction completely changes the equation. That is,

$y={\dfrac{3x+1}{(4x-1)^\frac{1}{2}}} \neq {\dfrac{(3x+1)^2}{\left((4x-1)^\frac{1}{2}\right)^2}}$

Edit 1: If you did want to square both top and bottom, not that you would have to square both sides:

$y^2= \left( {\dfrac{(3x+1)}{(4x-1)^\frac{1}{2}}} \right) ^2 = {\dfrac{(3x+1)^2}{\left((4x-1)^\frac{1}{2}\right)^2}}$

Edit 2: If your main concern is differentiating, I really encourage you to follow Nilabro's advice in your comments -- logarithmic differentiation. We would have

$y={\dfrac{3x+1}{(4x-1)^\frac{1}{2}}}$

$\ln (y)= \ln \left( {\dfrac{3x+1}{(4x-1)^\frac{1}{2}}} \right)$

which by using the properties of logarithms results in

$\ln (y)= \ln (3x+1) - \dfrac{1}{2} \ln (4x-1)$

then differentiating and solving for $y'$ gives

$\dfrac{1}{y} \ y'= \dfrac{3}{3x+1} - \dfrac{1}{2} \left( \dfrac{4}{4x-1} \right)$

$ y'= y \left( \dfrac{3}{3x+1} - \dfrac{1}{2} \left( \dfrac{4}{4x-1} \right) \right)$

$ y'= \left( {\dfrac{3x+1}{(4x-1)^\frac{1}{2}}} \right) \left( \dfrac{3}{3x+1} - \dfrac{1}{2} \left( \dfrac{4}{4x-1} \right) \right)$

$y' = \dfrac{6x-5}{(4x-1)^\frac{3}{2}},$

which is the same result randomgirl came to.

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You could square both sides but the problem turns into implicit problem instead of an explicit one.

$y=\frac{3x+1}{(4x-1)^\frac{1}{2}} \\\ $

We can square both sides:

$y^2=\frac{(3x+1)^2}{4x-1}$

Differentiate both sides:

$2yy'=\frac{[(3x+1)^2]'(4x-1)-[(4x-1)]'(3x+1)^2}{(4x-1)^2} \\\ $

$2 yy'=\frac{[2(3x+1) \cdot 3](4x-1)-[4](3x+1)^2}{(4x-1)^2} \\\ $

$2 yy'=\frac{6 (3x+1)(4x-1)-4(3x+1)^2}{(4x-1)^2} \\\ $

$2 yy'=(3x+1) \frac{6(4x-1)-4(3x+1)}{(4x-1)^2} \\$

Divide both sides by $2y$

$y'=\frac{3x+1}{2y} \frac{24x-6-12x-4}{(4x-1)^2} \\$

$y'=\frac{3x+1}{2y} \frac{12x-10}{(4x-1)^2} \\$

$y'=\frac{3x+1}{2} \frac{1}{y} \frac{12x-10}{(4x-1)^2} \\$

$y'=\frac{3x+1}{2} \frac{(4x-1)^\frac{1}{2}}{3x+1} \frac{12x-10}{(4x-1)^2} \\$

$y'=\frac{(4x-1)^\frac{1}{2}}{2} \frac{2(6x-5)}{(4x-1)^2} \\$

$y'=(4x-1)^\frac{1}{2} \frac{6x-5}{(4x-1)^2} \\$

$y'=\frac{6x-5}{(4x-1)^\frac{3}{2}}$