Question

find the derivative of y with respect to x

y=ln((1+sqrtx)/(x^5))

Answer 1

quotient rule

Answer 2

\[\ln\left(\frac{1+\sqrt{x}}{x^{5}}\right) \]?

Answer 3

quotient rule ..... after chain rule

Answer 4

yes mimi_x3

Answer 5

I would avoid the quotient rule by rewriting the argument of the log

Answer 6

nice

Answer 7

\[\ln(x^{-5}+x^{-9/5})\]who need to make life harder?
;)

Answer 8

i thought you meant:
\[y=\ln \left( 1+\sqrt x \over x^5 \right)=\Large \ln \left( 1+\sqrt x \right)-5\ln \left( x \right)\]

Answer 9

oh that's even better :)

Answer 10

The way paxpolaris did it was an easy solve but it does not lead to one of the answers in the multiple choices.

Answer 11

rationaliz denominator .... and maybe you need to combine the 2 fractions

Answer 12

\[\large y\ '={1 \over 1+ \sqrt x}\times \frac 12 \cdot \frac 1 {\sqrt x}-\frac 5x\]

Answer 13

what i ended up with was \[y=(1/(2x ^{1/2}(1+\sqrt{x})))-(5/x)\] if that is right... but its not an option on the homework choices.

Answer 14

yes that's right ...

is the denominator in at least one of you choices
2x(x-1)

Answer 15

\[\implies \Large y\ ' ={1 \over 2 \left(x+\sqrt x\right)} - \frac 5x\]

Answer 16

rationalize denominator:
\[\Large ={1\over 2 \left( x+ \sqrt x \right)}\cdot{\left( x- \sqrt x \right) \over\left( x- \sqrt x \right)} - \frac 5x\]
\[\Large ={x-\sqrt x \over 2\left( x^2-x \right)}-\frac 5x\]

Answer 17

\[\Large ={\left( x-\sqrt x \right)-10\left( x-1 \right) \over 2x(x-1)}\]

Answer 18

The options given are
\[10-9\sqrt{x}/2x(1+\sqrt{x})\]
\[-10-9\sqrt{x}/2x(1+\sqrt{x})\]
\[-10-9\sqrt{x}/2(1+\sqrt{x})\]
\[-10-9\sqrt{x}/2x\]

Answer 19

all right:\[\Large y\ ' ={1 \over 2 \sqrt x\left(1+\sqrt x\right)} - \frac 5x\]\[\Large ={\sqrt x \over 2 x\left(1+\sqrt x\right)} - {5\cdot2\left( 1+\sqrt x \right) \over 2 x\left(1+\sqrt x\right)}\]

Answer 20

\[\huge = {-10-9\sqrt x \over 2x \left( 1+\sqrt x \right)}\]

Answer 21

Thank you very much everyone. PaxPolaris special thanks!