How do you differentiate $y = {ln [\frac{x}{2 x + 3}]}^{\frac{1}{2}}$ $y=ln[x/(2x+3)]^(1/2)$ ?

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Answer 1 · 2017-07-22T16:41:09

Differentiate: $y = {ln [\frac{x}{2 x + 3}]}^{\frac{1}{2}}$ $y=ln[x/(2x+3)]^(1/2)$

Use the property $ln (a^{c}) = (c) ln (a)$ $ln(a^c) = (c)ln(a)$

$y = \frac{1}{2} ln (\frac{x}{2 x + 3})$ $y=1/2ln(x/(2x+3))$

Use the property $ln (\frac{a}{b}) = ln (a) - ln (b)$ $ln(a/b) = ln(a)-ln(b)$

$y = \frac{1}{2} ln (x) - \frac{1}{2} ln (2 x + 3)$ $y=1/2ln(x)-1/2ln(2x+3)$

Now, we may differentiate each term:

$\frac{d y}{d x} = \frac{1}{2 x} - \frac{1}{2} \frac{2}{2 x + 3}$ $dy/dx=1/(2x)-1/2 2/(2x+3)$

$\frac{d y}{d x} = \frac{1}{2 x} - \frac{1}{2 x + 3}$ $dy/dx=1/(2x)-1/(2x+3)$

Answer 2 · 2017-07-22T17:07:30

$\frac{d y}{d x} = \frac{3}{2 x (x + 3)} .$ $dy/dx=3/{2x(x+3)}.$

We have, $y = {ln {\frac{x}{2 x + 3}}}^{\frac{1}{2}} .$ $y=ln{x/(2x+3)}^(1/2).$

Using the familiar rules of Log. Fun., we get,

$y = \frac{1}{2} ln (\frac{x}{2 x + 3}) = \frac{1}{2} {ln x - ln (2 x + 3)} .$ $y=1/2ln(x/(2x+3))=1/2{lnx-ln(2x+3)}.$

$:. dy/dx=d/dx[1/2{lnx-ln(2x+3)}].$

$=1/2d/dx{lnx-ln(2x+3)},$

$=1/2[d/dx{lnx}-d/dx{ln(2x+3)}],$

$=1/2[1/x-1/(2x+3)d/dx{(2x+3)}],$

$=1/2[1/x-2/(2x+3)],$

$=1/2[{(2x+3)-2x}/{x(2x+3)}],$

$rArr dy/dx=3/{2x(2x+3)}.$

How do you differentiate y=ln[x/(2x+3)]^(1/2)y=ln[x2x+3]12y=ln[x/(2x+3)]^(1/2)?