1. How would one solve this;

where y is a function of x??

would appreciate any help.

2.

3. i see 1 equation with 2 unknowns.

what do we have to solve?

4. oops my bad, its alright i know how to solve it now. You are supposed to get

5. Originally Posted by Heinsbergrelatz
oops my bad, its alright i know how to solve it now. You are supposed to get

Product rule or chain rule, one or the other, look em up I forget them.

Maybe both who knows?

6. so x= y sin y^2

so for x-siny^2 dx/dy = 2ycos y^2

So dx/dy = sin y^2 + y2y cos y^2

= sin y^2 + 2y^2 cos y^2

thus dy/dy=

................ 1
= ---------------------

sin y^2 + 2y^2 cos y^2

7. Originally Posted by smokey
so x= y sin y^2

so for x-siny^2 dx/dy = 2ycos y^2

So dx/dy = sin y^2 + y2y cos y^2

= sin y^2 + 2y^2 cos y^2

thus dy/dy=

................ 1
= ---------------------

sin y^2 + 2y^2 cos y^2

no.

8. Originally Posted by Arcane_Mathematician
Originally Posted by smokey
so x= y sin y^2

so for x-siny^2 dx/dy = 2ycos y^2

So dx/dy = sin y^2 + y2y cos y^2

= sin y^2 + 2y^2 cos y^2

thus dy/dy=

................ 1
= ---------------------

sin y^2 + 2y^2 cos y^2

no.

That is dx/dy I believe not dy/dx
Bit of a trick really.

9. we have to express only in terms of y and x, sorry for not noting all the details. so its supposed to be;

10. That looks a bit ropey to me.

11. That looks a bit ropey to me.

well that is the only possible case when expressing the derivative in terms of x and y, got any other alternatives if you say its "ropey"?

12. Originally Posted by smokey
That looks a bit ropey to me.
He's absolutely correct.

13. Originally Posted by Arcane_Mathematician
Originally Posted by smokey
That looks a bit ropey to me.
He's absolutely correct.

Who me?

Yes he is correct, I did not notice what he did initially but looking again
I see he substituted back in the initial equation into the result to get a