Question

calculate the derivative with respect to x of y + 1/y=x^2+x

Answer 1

I know you have to take the derivative of both sides

Answer 2

not quite where to go from here

Answer 3

simply take the derivatice of y as you would for x, except add a y', since y is a function of x, you must include that.

for example: \[\frac{ d }{ dx }(y^2) = 2yy' = 2y \frac{ dy }{ dx }\]

Answer 4

Then, you can factor out a dy/dx and solve for dy/dx

Answer 5

k. can u go throgh the steps please. Y^2 ?

Answer 6

\[y' - \frac{ 1 }{ y^2 }y'\] is the left side. Now, do the right side. group like terms (remember from algebra). factor out and solve for y'.
\[(ax+ay) = a(x+y)\]

Answer 7

for \[\frac{ d }{ dx }(a^n) = na^{n-1} \] remember

Answer 8

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

Answer 9

\[\frac{ 1 }{ y } = y^{-1}\]

Answer 10

want to make sure the equations was inputted correctly

Answer 11

So, take the derivative as you would for the whole thing.

Answer 12

except whenever u take derivative of "y" in this, you multiply it by y'

Answer 13

the derivative is to tell you that it's just a derivative with respect to x. Kind of like chain rule.

Answer 14

k

Answer 15

is the first step d / dx (y) + d / dx (1/y) = d / dx ( x^2 + x )

Answer 16

\[\frac{ dy }{ dx }=\frac{ 2x+1 }{1-\frac{ 1 }{ y^2 } }\]

Answer 17

can you show me the first step; just getting used to using the equation button

Answer 18

i know you have to take the derivative of both sides
d/ dx (y) + d/dx ( 1 / y ) = d / dx (x^2 + x)

Answer 19

then do I write in the y' on each side

Answer 20

okay let me write it in full

Answer 21

thanks

Answer 22

okay let e solve it on my book then i am gonna attach it as a pic coz it takes time having to do it here

Answer 23

yes it does; thanks

Answer 24

I am trying another while waiting for the pic

Answer 25

\[\sin \left( x+y \right)=\left( x+\cos y \right)\]

Answer 26

Will I see the pic in this screen

Answer 27

yes i guess

Answer 28

did you already post the answer in full

Answer 29

can u see it?

Answer 30

yes. I will try to open it

Answer 31

where did the 1 in the 1 - 1 / y squared come from

Answer 32

do I have to write the y squared beside the (2x + 1)

Answer 33

its like x-xy then you can write x(1-y) so thats how i got the 1

Answer 34

you can simplify it according to how you understand it

Answer 35

k

Answer 36

can you go through another one; I already posted it: sin ( x + y) = (x + cos y)

Answer 37

I started with \[\frac{ d }{ dx }\sin x)+ \]

Answer 38

d / dx (sin x) + d/dx(sin y) = d/dx (x) + d/dx (cos y)

Answer 39

okay i`m on it

Answer 40

Great

Answer 41

\[\frac{ dy }{ dy }=\frac{ 1-\cos(x) }{ 1+\sin(y) }\]
thats what i got what did u get

Answer 42

\[y`=(\cos x + y ) = 1 - \sin y)\]

Answer 43

oh sorry my mistake

Answer 44

I know I missed a step

Answer 45

i forgot to differentiate the inner function

Answer 46

the book`s answer has 1- cos(x+y) / cos ( x + y) + sin y

Answer 47

not sure how to get to the books answer from what your anser and my answer

Answer 48

could you verify how to differentiate the inner function

Answer 49

dy/dx=sin(x+y) Put x+y=u then,differentiate

Answer 50

yeah i am on it

Answer 51

Great- you are a big help!

Answer 52

its a mind challenging but i never give up i am stil coming with a solution not long

Answer 53

Thanks!

Answer 54

got it

Answer 55

\[\frac{ d(\sin(x+y)) }{ dx }=\cos(x+y)(1+\frac{ dy }{ dx })\]
so we can take it from here

Answer 56

\[\frac{ dy }{dx }=\frac{ 1-\cos(x+y) }{ \sin(y)+\cos(x+y)) }\]
thats what i got check it tell me is i got it wrong even now

Answer 57

that`s exactly what is in the answer key of the book

Answer 58

what was your second step; before the answer

Answer 59

something like that