Question

dy/dx = √ (xy+1). Would the answer be y/2√(xy+1)?

Answer 1

\[\frac{ y }{ 2\sqrt{xy+1} }\]

Answer 2

is that a differential equation ?

or u just wanto find derivative of : \(y = \sqrt{xy+1}\) ?

Answer 3

implicit differentiation

Answer 4

I think just needs the derivative but you're missing a -x in the numerator y/2square root(xy+1)-x

Answer 5

denominator*

Answer 6

Alright, whats the actual input equation ?


\(y = \sqrt{xy+1}\)


or


\(\frac{dy}{dx} = \sqrt{xy+1}\)

?

Answer 7

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

Answer 8

if it is second one, you need to find second derivative i guess...
could you pls clarify ha

Answer 9

I'm so confused now >.<

Answer 10

what??..sorru.

Answer 11

Is your original question y = square root (xy+1) ? or is dy/dx

Answer 12

dy/dx....

Answer 13

wait no,it's said x= square root (xy+1)

Answer 14

and then find dy/dx

Answer 15

Then your answer is correct.

Answer 16

\(\large x = \sqrt{xy+1}\)


differentiate bobth sides with.respect.to x


\(\large 1 = \frac{1}{2\sqrt{xy+1}} (xy+1)'\)

Answer 17

^^that comes by chain rule
differentiate the thing inside parenthesis, you wud get some dy/dx term, solve for it

Answer 18

how would u do that?..

Answer 19

Product rule

Answer 20

nevermine i got it. Thanks!

Answer 21

\(\large x = \sqrt{xy+1}\)


differentiate bobth sides with.respect.to x


\(\large 1 = \frac{1}{2\sqrt{xy+1}} (xy+1)'\)

\(\large 1 = \frac{1}{2\sqrt{xy+1}} \left((xy)' + (1)'\right) \)

\(\large 1 = \frac{1}{2\sqrt{xy+1}} \left((xy)' + 0\right) \)

\(\large 1 = \frac{1}{2\sqrt{xy+1}} \left((xy)' \right) \)

Answer 22

apply product rule for \((xy)'\)