Question

Question #b7391

Answer 1

You use the chain rule for the right side:

`(d(cos(piy)))/dx = -(d(cos(piy)))/dy dy/dx`

The entire derivative is:

`dy/dx= (-3x^2y^2)/(2x^3y+ pisin(piy))`

Explanation:

Given: `x^3y^2=cos(piy)`

Write the equation so that it is equal to zero:

`x^3y^2-cos(piy)= 0`

Differentiate each term:

`(d(x^3y^2))/dx-(d(cos(piy)))/dx= 0" [1]"`

For the first term we use the product rule:

`(d(x^3y^2))/dx = (d(x^3))/dxy^2+x^3(d(y^2))/dx`

The first term on the right is trivial:

`(d(x^3y^2))/dx = 3x^2y^2+x^3(d(y^2))/dx`

The second term on the right requires the chain rule:

`(d(x^3y^2))/dx = 3x^2y^2+x^3(d(y^2))/dy dy/dx`

`(d(x^3y^2))/dx = 3x^2y^2+2x^3y dy/dx`

Substitute into equation [1]:

`3x^2y^2+2x^3y dy/dx-(d(cos(piy)))/dx= 0`

The last term requires the use of the chain rule:

`3x^2y^2+2x^3y dy/dx-(d(cos(piy)))/dy dy/dx= 0`

When we differentiate the cosine function with respect to y, we must remember to multiply by `pi`

`3x^2y^2+2x^3y dy/dx+ pisin(piy) dy/dx= 0`

Subtract `3x^2y^2` from both sides:

`2x^3y dy/dx+ pisin(piy) dy/dx= -3x^2y^2`

Remove a common factor of `dy/dx`:

`(2x^3y+ pisin(piy))dy/dx= -3x^2y^2`

Divide both sides by the leading coefficient:

`dy/dx= (-3x^2y^2)/(2x^3y+ pisin(piy))`