Question

Hey everyone i am struggling to finish of my equation.
Differentiate with respect to the x: y= ln2x / Sqrt x.

I have this so far: Quotient rule = v du/dx - u dv/dx / v^2.
u = ln2x
v = x ^1/2
du/dx = 1/x
dv/dx= 1/2 x^1/2
(x^1/2) 1/x - (ln2x) 1/2 x^1/2 / x^1/2^2.

Answer 1

also i know the answer should be: dy/dx (1-1/2 ln2x) / sqrt x^2.

Answer 2

\[y=\frac{\ln2x}{\sqrt x}\]
\[\begin{align*}y'&=\frac{x^{1/2}\left(\dfrac{1}{x}\right)-\ln2x\left(\dfrac{1}{2}x^{-3/2}\right)}{\left(x^{1/2}\right)^2}\\\\
&=\frac{x^{-1/2}-\dfrac{1}{2}x^{-3/2}\ln2x}{x}\\\\
&=x^{-3/2}-\frac{1}{2}x^{-5/2}\ln2x\\\\
&=x^{-5/2}\left(x^{2/2}-\frac{1}{2}x^{0/2}\ln2x\right)\\\\
&=x^{-5/2}\left(x-\frac{1}{2}\ln2x\right)\\\\
&=x^{-5/2}\left(x-\ln\sqrt{2x}\right)
\end{align*}\]
That's where I would stop, but you could definitely continue with some log property manipulation. I don't see how one would obtain what you say the solution should be.

Answer 3

@SithsAndGiggles in the first line of the derivative of the numenator the derivative of the squareroot of x isn't x^(-3/2) but ^(-1/2). When you take the derivative in a rational function in the numenator you treat it having a positive exponent.

Answer 4

\[y'=\frac{ x ^{\frac{ 1 }{ 2 }} \left( \frac{ 1 }{ x } \right) - \ln 2x \left( \frac{ 1 }{ 2 } x ^{\frac{ -1 }{ 2 }} \right) }{ x }\]
\[=\frac{ x \left( \frac{ 1 }{ x } \right) - \ln 2x \left( \frac{ 1 }{ 2 } \right) }{ x ^{\frac{ 3 }{ 2 }} }\]
\[=\frac{ 1 - \frac{ 1 }{ 2 }\ln (2x)\ }{ x ^{\frac{ 3 }{ 2 }} }\]

Answer 5

@sweetsunray yes you're right, sorry. I must have thought \(\dfrac{1}{\sqrt x}\) was in the denominator as opposed to \(\sqrt x\).