Question #b7391

Question

1107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-24T17:19:49

You use the chain rule for the right side:

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

The entire derivative is:

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

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:

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