Performing implicit differentiation involves differentiating each term on its own. We assume that yy is a function of xx so that any time we differentiate a function of yy, we will still have our unknown dy/dxdydx term left over.
d/dx 2 = color(red)(d/dx (e^(xy^3 - x))) - color(green)(d/dx (y^2x^3)) + d/dx (y)ddx2=ddx(exy3−x)−ddx(y2x3)+ddx(y)
The first term is a constant, therefore the derivative is 00. The last term is simply dy/dxdydx. For clarity, lets go over the other two terms individually.
For the color(red)"red"red term, we can use the fact that d/dx(e^f(x)) = e^f(x)f'(x).
color(red)(d/dx (e^(xy^3 - x)) = (e^(xy^3 - x)) d/dx (xy^3 - x) )
color(red)(=(e^(xy^3 - x)) (d/dx(xy^3)-d/dx(x)))
Apply the product rule, d/dx f(x)g(x) = f'(x)g(x)+f(x)g'(x).
color(red)(=(e^(xy^3 - x))(d/dx(x)*y^3 + x*d/dx(y^3) - 1))
color(red)(=(e^(xy^3 - x))(y^3 + 3xy^2 dy/dx - 1))
The green term is easier, we just need to apply the product rule again.
color(green)(d/dx(y^2x^3) = d/dx(y^2)*x^3 + y^2 d/dx(x^3))
color(green)(=2y dy/dx *x^3 + 3y^2x^2)
Now we can put everything back together. Don't forget that the middle term was negative.
0 = e^((xy^3 - x))(y^3 + 3xy^2 dy/dx - 1) - 2y dy/dx *x^3 - 3y^2x^2 +dy/dx
Separate the terms that have dy/dx and move them to the left hand side.
-e^((xy^3 - x))3xy^2 dy/dx+2x^3y dy/dx-dy/dx = e^((xy^3 - x))(y^3-1)- 3y^2x^2
(-e^((xy^3 - x))3xy^2+2x^3y -1)dy/dx = e^((xy^3 - x))(y^3-1)- 3y^2x^2
Solve for dy/dx.
dy/dx = -(e^((xy^3 - x))(y^3-1)- 3y^2x^2)/(e^((xy^3 - x))3xy^2+2x^3y -1)
