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.

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

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

.

 

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

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