How do you implicitly differentiate #2= e^(xy^2-x^3y)-y^2x^3+y #?

Question

How do you implicitly differentiate #2= e^(xy^2-x^3y)-y^2x^3+y     #?

Luke Alvarez · Accepted Answer

#y' = -(y(3x^2ye^(x^3y)+3xe^(xy^2)-ye^(xy^2)))/(2x^3ye^(x^3y)-e^(x^3y)-2xye^(xy^2)+x^3e^(xy^2)#

Given: #2 = e^(xy^2-x^3y) - y^2x^3 + y#

Derivative rules: #(k)' = 0# #(e^u)' = u'e^u# Product rule: #(uv)' = uv' + vu'#

Use the product rule twice for #e#: Let #u = xy^2 - x^3y#

#u' = 2xyy' + y^2 -x^3y' -3x^2y#

#(e^u)' = (2xyy' + y^2 -x^3y' -3x^2y)e^(xy^2-x^3y)#

#(-y^2x^3)' = -3x^2y^2 -2x^3yy'#

Put it all together #0 = (2xyy' + y^2 -x^3y' -3x^2y)e^(xy^2-x^3y) -3x^2y^2 - 2x^3yy' + y'#

Distribute the #e#:

#0 = 2xyy'e^(xy^2-x^3y) + y^2e^(xy^2-x^3y)-x^3y'e^(xy^2-x^3y) -3x^2y e^(xy^2-x^3y) -3x^2y^2 - 2x^3yy' + y'#

Move all the #y'# terms to the left side: #-2xyy'e^(xy^2-x^3y) + x^3y'e^(xy^2-x^3y) +2x^3yy' - y' = y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2#

Factor the #y'#: #y'(-2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y) +2x^3y - 1) = y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2#

Divide:

#y' = (y^2e^(xy^2-x^3y)-3x^2y e^(xy^2-x^3y) -3x^2y^2)/((-2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y) +2x^3y - 1))#

Factor out a negative from the numerator and rearrange: #y' = -(3x^2y^2 + 3x^2y e^(xy^2-x^3y) -y^2e^(xy^2-x^3y))/(2x^3y - 1 -2xye^(xy^2-x^3y) + x^3e^(xy^2-x^3y))#

Make #e^(xy^2-x^3y) = e^(xy^2)e^(-x^3y)#

#y' = -(3x^2y^2 + 3x^2y e^(xy^2)e^(-x^3y) -y^2e^(xy^2)e^(-x^3y))/(2x^3y - 1 -2xye^(xy^2)e^(-x^3y) + x^3e^(xy^2)e^(-x^3y))#

Factor #ye^(-x^3y)# from both numerator and #e^(-x^3y)# from the denominator. Realize that #e^(-x^3y)e^(x^3y) = e^0 = 1#

#y' = -(ye^(-x^3y)(3x^2ye^(x^3y) + 3xe^(xy^2) - ye^(xy^2)))/(e^(-x^3y)(2x^3ye^(x^3y) - e^(x^3y) - 2xye^(xy^2)+x^3e^(xy^2)))#

Cancel common factors: #y' = -(y(3x^2ye^(x^3y)+3xe^(xy^2)-ye^(xy^2)))/(2x^3ye^(x^3y)-e^(x^3y)-2xye^(xy^2)+x^3e^(xy^2)#

Emma Aldridge · Answer

undefined To implicitly differentiate the equation $ 2 = e^{xy^2-x^3y} - y^2x^3 + y $ with respect to $ x $, you will differentiate each term individually while applying the chain rule and product rule where necessary.

Differentiating the first term $ e^{xy^2-x^3y} $ with respect to $ x $ using the chain rule:
$$
\frac{d}{dx} (e^{xy^2-x^3y}) = \frac{d}{dx}(xy^2-x^3y) \cdot e^{xy^2-x^3y}
$$

Differentiating $ xy^2-x^3y $ with respect to $ x $:
$$
\frac{d}{dx}(xy^2-x^3y) = y^2 - 3x^2y
$$

Therefore, the first term becomes:
$$
(e^{xy^2-x^3y}) \cdot (y^2 - 3x^2y)
$$

Differentiating the second term $ -y^2x^3 $ with respect to $ x $:
$$
\frac{d}{dx} (-y^2x^3) = -3y^2x^2
$$

Differentiating the third term $ +y $ with respect to $ x $:
$$
\frac{d}{dx}(y) = \frac{dy}{dx}
$$

So, the derivative of the equation with respect to $ x $ is:
$$
0 = (e^{xy^2-x^3y}) \cdot (y^2 - 3x^2y) - 3y^2x^2 + \frac{dy}{dx}
$$

This is the implicit differentiation of the given equation with respect to $ x $.