Implicit Differentiation of y*sin(x^2) = x*sin(y^2)

[Prim3]

Hello, I'm having a bit of trouble solving some questions regarding Implicit Differentiaton:

y*sin(x^2) = x*sin(y^2)

Don't know how to start. Should I use the product rule? If so, how?

Thanks.

 

[stapel]

y*sin(x^2) = x*sin(y^2)

Don't know how to start. Should I use the product rule? If so, how?

Would you know how to differentiate x sin(x^2)...?

Eliz.

 

[Prim3]

Re: Implicit Differentiation

Well in my opinion:

We can use the product rule and get:

x * dy/dx (sin(x^2)) + (sin(x^2)) * dy/dx (x)

Then we get:

x * dy/dx (sin(x^2)) + (sin(x^2))

Chain rule then, so:

x * cos(x^2) * dy/dx (x^2) + (sin(x^2))

x * cos(x^2) * (2x) + (sin(x^2))

2cos(x^2)(x^2) + (sin(x^2))

Correct?

Thanks.

 

[Deleted member 4993]

Re: Implicit Differentiation

Well in my opinion:

We can use the product rule and get:

x * dy/dx (sin(x^2)) + (sin(x^2)) * dy/dx (x) <<<< It should be written as

x * d/dx (sin(x^2)) + (sin(x^2)) * d/dx (x) <<<<<< no 'y' here - you did not have a 'y' to begin with

= x * d/dx (sin(x^2)) + (sin(x^2)) <<<<<< no 'y' here - you did not have a 'y' to begin with

=x * cos(x^2) * d/dx (x^2) + (sin(x^2))<<<<<< no 'y' here - you did not have a 'y' to begin with

=x * cos(x^2) * 2* x + (sin(x^2))

=2x^2 * cos(x^2) + (sin(x^2))

Then we get:

x * dy/dx (sin(x^2)) + (sin(x^2))

Chain rule then, so:

x * cos(x^2) * dy/dx (x^2) + (sin(x^2))

x * cos(x^2) * (2x) + (sin(x^2))

2cos(x^2)(x^2) + (sin(x^2))

Correct?

Thanks.

 

[stapel]

Well in my opinion: We can use the product rule and get:

x * dy/dx (sin(x^2)) + (sin(x^2)) * dy/dx (x)

x * dy/dx (sin(x^2)) + (sin(x^2))

x * cos(x^2) * dy/dx (x^2) + (sin(x^2))

x * cos(x^2) * (2x) + (sin(x^2))

2cos(x^2)(x^2) + (sin(x^2))

Yes. Now do the exact same process, but with a "y" in place of the "x" in front of the sine. Since (dy/dx)(y) is just dy/dx, you'll end up with a "dy/dx" where here you'd had a "1".

Then follow the same process on the right-hand side of the equation.

Then solve for dy/dx, and you're done!

Eliz.