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

Question

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

Scarlett Allison · Accepted Answer

#y'=(e^(xy^3-x(1-y^3)+y^2)/(3xy^2e^(xy^3-x)+2xy))# Differentiating with respect to #x# #0=e^(xy^3-x)(y^3+x*3y^2y'-1)-y^2-x*2y*y'# this gives #0=e^(xy^3-x)(y^3-1)-y^2+y'(3xy^2e^(xy^3-x)+2xy)#

Paisley Johnson · Answer

undefined To implicitly differentiate the equation $2 = e^{xy^3 - x} - y^2x$, we differentiate both sides of the equation with respect to $x$, treating $y$ as an implicit function of $x$ using the chain rule and the product rule where necessary.

1. Differentiate $e^{xy^3 - x}$ with respect to $x$ using the chain rule:
$$ \frac{d}{dx} \left( e^{xy^3 - x} \right) = e^{xy^3 - x} \cdot (y^3 + (-1)) \cdot y' = e^{xy^3 - x} \cdot (y^3 - 1) \cdot y' $$

2. Differentiate $y^2x$ with respect to $x$ using the product rule:
$$ \frac{d}{dx} (y^2x) = y^2 \cdot 1 + 2y \cdot y' $$

3. Set up the equation for the implicit differentiation:
$$ 0 = e^{xy^3 - x} \cdot (y^3 - 1) \cdot y' - y^2 + 2yy' $$

4. Solve for $y'$:
$$ y' (e^{xy^3 - x} \cdot (y^3 - 1) + 2y) = y^2 - e^{xy^3 - x} \cdot (y^3 - 1) $$

$$ y' = \frac{y^2 - e^{xy^3 - x} \cdot (y^3 - 1)}{e^{xy^3 - x} \cdot (y^3 - 1) + 2y} $$