Question

Find dy/dx if y= square root 2x^2+1 + 3/squareroot3x+1

Answer 1

\[\frac{dy}{dx} = \frac{2x}{\sqrt{2x^{2}+1}}-\frac{9}{2(3x+1)\sqrt{3x+1}}\]

Answer 2

Possible steps here http://www.wolframalpha.com/input/?i=square+root+%282x^2%2B1%29+%2B+3%2F%28squareroot+%283x%2B1%29%29

Answer 3

\[y = \sqrt{2x^2 +1} +\frac{3}{\sqrt{3x+1}}\]

Answer 4

yes

Answer 5

2x^2+1 power 1/2 + 3/squareroot3x+1 then dy over dx???

Answer 6

Oh

Answer 7

why you deleted -.-

Answer 8

\[\frac{dy}{dx} = \frac{1}{2}*4x*\frac{1}{\sqrt{2x^2-1}} - \frac{3*3}{(3x+1)^{3/2}}\]

Answer 9

i forgot 3 and - sign

Answer 10

yeaaa

Answer 11

and i forgot another thing multiply by 1/2 to the second expression (3x+1) one

Answer 12

okay take your time :)

Answer 13

\[\frac{dy}{dx} = \frac {1}{2}*(2x^2+1)^{1/2 -1}*{4x} +3*{\frac{-1}{2}(3x +1 )^{-1/2-1} *3}\]

Answer 14

the powers are 1/2 -1 and -1/2-1

Answer 15

\[\large{\frac{dy}{dx} = \frac{4x}{2{\sqrt{2x^2+1}}} - \frac{9}{2{(3x+1)^{\frac{3}{2}}}}}\]

Answer 16

this should be the answer

Answer 17

let me attach something it might help you better