Given y=sin(x^2 + y).
Find dy/dx by implicit differentiation.

Question

Given y=sin(x^2 + y).
Find dy/dx by implicit differentiation.

mathstudent55 · Answer

$y = \sin(x^2 + y) $

$y' = (2x + y')\cos(x^2 + y)$

$y' = 2x\cos(x^2+y)+y'\cos(x^2+y)$

myininaya · Answer

Looks good so far.

mathstudent55 · Answer

$y' - y'\cos(x^2+y) = 2x\cos(x^2 +y)$

$y'[1 -\cos(x^2+y)] = 2x\cos(x^2 +y)$

mathstudent55 · Answer

$y' = \dfrac{ 2x\cos(x^2 +y)  }{ 1 -\cos(x^2+y)  }$

myininaya · Answer

1-cos^2(u) can be written as sin^2(u) if you wanted

anonymous · Answer

Do you need the work checked?

mathstudent55 · Answer

Great, thanks.

carson889 · Answer

It is correct to the best of my knowledge

myininaya · Answer

also sin^2(x^2+y) on bottom

myininaya · Answer

and if you wanted you could write sin^2(x^2+y) as y^2
since y=sin(x^2+y)
just if you wanted

myininaya · Answer

$$1-\cos^2(x^2+y)=\sin^2(x^2+y) =y^2$$

myininaya · Answer

oh wait there is no square on that cosine on bottom
don't listen to me

mathstudent55 · Answer

Here we had 1 - cos u in the denom, not 1 - cos^2 u, so it's not sin^2 u in the denom, right?

asnaseer · Answer

@myininaya - I don't see a $\cos^2$ term in the original answer that mathstudent55 wrote down which was:$$y' = \dfrac{ 2x\cos(x^2 +y)  }{ 1 -\cos(x^2+y)  }$$

myininaya · Answer

yes yes yes i said don't listen to me already

myininaya · Answer

lol

mathstudent55 · Answer

I'll leave the answer as this, then:
$y' = \dfrac{ 2x\cos(x^2 +y)  }{ 1 -\cos(x^2+y)  }$

asnaseer · Answer

:)

myininaya · Answer

yes mathstudent is fine as is

mathstudent55 · Answer

Thanks to everyone for your help!

anonymous · Answer

How did you get 1-cos(x^2 + y) in the denom? I get stuck on that part.

myininaya · Answer

rjaza are you good with 
$$y' - y'\cos(x^2+y) = 2x\cos(x^2 +y)  $$

myininaya · Answer

i mean do you understand up until that point?

anonymous · Answer

Yes I got that part I subtracted dy/dx cos(x^2 +y) from both sides

anonymous · Answer

just don't understand where 1-cos comes from ?

myininaya · Answer

ok so if you factor the y' out on the left hand side then you have
$$y'(1-\cos(x^2+y))=2x \cos(x^2+y)$$

myininaya · Answer

to solve that for y' you need to divide both sides by (1-cos(x^2+y))

anonymous · Answer

how does the -1 happen?

myininaya · Answer

-1?

anonymous · Answer

I did factor out dy/dx

anonymous · Answer

yes 1-cos(x^2 + y) where does the 1-cos come from. Sorry not -1

myininaya · Answer

$$\text{ factoring example } x-x^2=x(1-x)$$

myininaya · Answer

Both of these terms in the example have a common factor x

anonymous · Answer

Oh okay I see now!

myininaya · Answer

coolness

anonymous · Answer

thank you :)

myininaya · Answer

np