Question

derivative: (x-1)^(x)^1/2

Answer 1

\[(x-1)^{\sqrt{x}}\]

Answer 2

i took the log and got ln(x-1)\[\sqrt{x}\]

Answer 3

\[\ln(x-1)\sqrt{x}\]

Answer 4

differentiate using product rule.

Answer 5

Then i used the product rule and got\[(\sqrt{x}/(x-1)) + (\ln(x-1)/2\sqrt{x)}\]

Answer 6

is my answer correct?

Answer 7

yes

Answer 8

Also, how do you know when to use the quotient rule? sometimes given a fraction, we are asked to solve it with the product rule.

Answer 9

thank you :)

Answer 10

quotient rule is same as product rule u/v = uv^-1

Answer 11

and you are missing something.

Answer 12

oh ok and for limits when do we use the L'hospital's rule?

Answer 13

what am i missing?

Answer 14

L'hospital's is used when the limit you are evaluating is indeterminate, such as 4/0 or 0/0. You don't take the quotient rule with L'hospital' though, you take the derivative of the numerator and denominator separately

Answer 15

@brainshot3 thank you!

Answer 16

lny = x something, <---- took derivative here
1/y dy/dx = x <--- derivative of x's

dy/dx = y times derivative of x's

Answer 17

is this implicit differentiation? i don't really understand it

Answer 18

it's same as differentiation, but you will have two variables instead of one.

Answer 19

oh ok

Answer 20

check mit ocw video on single variable calculus.

Answer 21

so my answer should be \[lny=\sqrt{x}/x-1 + \ln(x-1)/2\sqrt{x}\]

Answer 22

ok thank you!

Answer 23

no your answer should be

y times the value of left. and put the value of y.

Answer 24

You want dy/dx alone on one side and you do that by multiplying out the y

Answer 25

\[dy/dx=y \sqrt{x}/x-1 + \ln(x-1)/2\sqrt{x} \]

Answer 26

Looks correct

Answer 27

because the derivative of ln y = 1/y

Answer 28

thank you all for your help :)

Answer 29

Wait, you forgot to distribute the y

Answer 30

y = (x-1)^x

so your answer should be
\[dy/dx = (x-1)^x (√x/(x−1)+ln(x−1)/2√x) \]

Answer 31

oh i got it now! thank you :)