Question

How do you find dy/dx in terms of x and y if (x^3)y-x-2y-6=0?

Answer 1

`(3x^(2)y+x^(3)y')-1-2y'=0`

Explanation:

For this, you use implicit differentiation (basically regular differentiation, but with y as well as x). It works the same way as regular differentiation, except every time you have to take the derivative of y, you have to put a `y'` after it.For the first term, you have to use the product rule, which is:

`(f'(x)×g(x))+(f(x)×g'(x))`

Here `f(x)=x^3` and `g(x)=y`. So you take the derivative of `x^3` and multiply by `y`, then then switch, so you get the first term (surrounded by parentheses in the answer).

The rest is pretty simple, the derivative of `x` is `1`, the derivative of `-2y` is `-2y'` and `6` and `0` are both `0`. You put all together, and you get:

`(3x^(2)y+x^(3)y')-1-2y'=0`

Which is the answer : )

Answer 2

`(3x^(2)y+x^(3)y')-1-2y'=0`

Explanation:

For this, you use implicit differentiation (basically regular differentiation, but with y as well as x). It works the same way as regular differentiation, except every time you have to take the derivative of y, you have to put a `y'` after it.For the first term, you have to use the product rule, which is:

`(f'(x)×g(x))+(f(x)×g'(x))`

Here `f(x)=x^3` and `g(x)=y`. So you take the derivative of `x^3` and multiply by `y`, then then switch, so you get the first term (surrounded by parentheses in the answer).

The rest is pretty simple, the derivative of `x` is `1`, the derivative of `-2y` is `-2y'` and `6` and `0` are both `0`. You put all together, and you get:

`(3x^(2)y+x^(3)y')-1-2y'=0`

Which is the answer : )