ln Diff Example - # 4

[Jason76]

$\displaystyle y = \sqrt{\dfrac{x-5}{x^{8} + 4}}$

$\displaystyle y = (\dfrac{x-5}{x^{8} + 4})^{1/2}$

$\displaystyle \ln(y) = \ln[(\dfrac{x-5}{x^{8} + 4})^{1/2}]$

$\displaystyle \ln(y) = \dfrac{1}{2}\ln[(\dfrac{x - 5}{x^{8} + 4})]$

$\displaystyle \ln(y) = \dfrac{1}{2}\ln[(x - 5)] - \dfrac{1}{2} \ln[(x^{8} + 4)]$

 

[Deleted member 4993]

$\displaystyle y = \sqrt{\dfrac{x-5}{x^{8} + 4}}$

$\displaystyle y = (\dfrac{x-5}{x^{8} + 4})^{1/2}$

$\displaystyle \ln(y) = \ln[(\dfrac{x-5}{x^{8} + 4})^{1/2}]$.............. why ln? ...............use product/quotient rule and chain rule.

$\displaystyle \ln(y) = \dfrac{1}{2}\ln[(\dfrac{x - 5}{x^{8} + 4})]$

$\displaystyle \ln(y) = \dfrac{1}{2}\ln[(x - 5)] - \dfrac{1}{2} \ln[(x^{8} + 4)]$

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[HallsofIvy]

While, as Suhotosh Kahn said, it is not necessary to use "logarithmic differentiation", you certainly can.

$\displaystyle y= \left(\dfrac{x- 5}{x^8+ 4}\right)^{1/2}$
$\displaystyle ln(y)= \ln\left(\dfrac{x- 5}{x^8+ 4}\right)^{1/2}= \dfrac{1}{2}ln\left(\dfrac{x- 5}{x^8+ 4}\right)$
$\displaystyle ln(y)= \dfrac{1}{2}ln(x- 5)- \dfrac{1}{2}ln(x^8+ 4)$

Now differentiate: $\displaystyle (ln(y))'= \dfrac{y'}{y}$, $\displaystyle (ln(x-5))'= \dfrac{1}{x- 5}$ and $\displaystyle (ln(x^8+ 4))'= \dfrac{8x^7}{x^8+ 4}$ (chain rule: derivative of $\displaystyle x^8+ 4$ is $\displaystyle 8x^7$)

So $\displaystyle \dfrac{y'}{y}= \dfrac{1}{2(x- 5)}- \dfrac{4x^7}{x^8+ 4}$\
$\displaystyle y'= y\left(\dfrac{1}{2(x- 5)}- \dfrac{4x^7}{x^8+ 4}\right)$

$\displaystyle y'= \sqrt{\dfrac{x- 5}{x^8+ 4}}\left(\dfrac{1}{2(x- 5)}- \dfrac{4x^7}{x^8+ 4}\right)$.