Question

Find h′(2) if h(x)=ln(x+sqrt(x^2−1√).

Answer 1

i got 0.3452 this is not right

Answer 2

\[h'=\frac{ 1- \frac{ 2x }{ x^2-1 } }{ x+\sqrt{x^2-1} }\]

Answer 3

i don't know how you got the numerator

Answer 4

waıt this\[1-\frac{ x }{ 2\sqrt{x^2-1} }\]

Answer 5

yo get thıs when you derıve
\[x+\sqrt{x^2-1}\]

Answer 6

\[h'(x)=\frac{ 1-\frac{ x }{ 2\sqrt{x^2-1}} }{ x+\sqrt{x^2-1} }\]

Answer 7

since\[y=\ln x\]
\[y'=\frac{ x' }{ x }\]

Answer 8

in my numerator i got
\[1+\frac{ 1 }{ 2\sqrt{x^2-1} }\]

Answer 9

how did you get a x where i have a 1?

Answer 10

you are correct but there ıs no 2,but x on top

Answer 11

where does the x come from?

Answer 12

chain rule for (x^2 - 1)

Answer 13

\[\sqrt{x ^{2}-1})'= 1/2(2x)(x ^{2}-1 )^{-1/2} =\frac{ x }{ \sqrt{x ^{2}-1} }\]

Answer 14

chain rule doesn't stop on the square root; it extends to the expression inside the square root.

Answer 15

ok that makes sense thanks for explaining it

Answer 16

going on, \[h\prime (2)\] equals what value?

Answer 17

0.57735

Answer 18

right. or \[h\prime (2) = \frac{ \sqrt{3} }{ 3 }\]