Question

I need to find dy/dx of the function in the first photo, the second photo is the result i'm supposed to get and the last photo is my result.

Answer 1

Explanation:

We begin with the function:
`4x^2y-x+y=1-sin(y^2)`

The first thing that we need to do is implicitly differentiate this function with respect to x.

Let us differentiate each piece individually and then combine them.

First,
`4x^2y`

There are two thing that we need to keep in mind with this: we will need to apply a product rule and when we differentiate a y with respect to x, we are left with a
`(dy)/(dx)`.

`4x^2y\\4[(2xy)+(x^2(dy)/(dx))}]\\8xy+4x^2(dy)/(dx)`

Now,
`-x\\-1`

And now,
`y\\\\(dy)/(dx)`

And lastly,
`1-sin(y^2)`

For this, we will need to apply the chain rule.

`1-sin(y^2)\\-(2y(dy)/(dx)) cos(y^2)\\-2y(dy)/(dx) cos(y^2)`

Now that we have differentiated each part of this function, we can combine them together to get
`8xy+4x^2(dy)/(dx) -1+(dy)/(dx) =-2y(dy)/(dx) cos(y^2)`

Our next course of action will be getting all of the terms with
`(dy)/(dx)` onto one side so that we can factor it out.

When this is done, we end up with
`4x^2(dy)/(dx) +(dy)/(dx) +2y(dy)/(dx) cos(y^2)=-8xy+1`

Once factored, we will get
`(dy)/(dx) (4x^2+1+2ycos(y^2))=-8xy+1\\\\(dy)/(dx)=(-8xy+1)/( 4x^2+1+2ycos(y^2))\\\\(dy)/(dx)=-(8xy-1)/( 4x^2+1+2ycos(y^2))`

We then were able to isolate
`(dy)/(dx)` and then factor out a -1 from the numerator to get our final answer.

Regarding your answer, From what I can see, you ended up incorrectly factoring out the
`(dy)/(dx)` from each side before you moved all of the terms to one side. This does not work as there are terms that do not have
`(dy)/(dx)` that are having it factored out of them.

Answer 2

I've attached the solution.