How do you implicitly differentiate #-y=x-sqrt(x^3y-y) #?

Question

How do you implicitly differentiate #-y=x-sqrt(x^3y-y)    #?

Charles Arocha · Accepted Answer

#y'=(2sqrt(x^3y-y)-3x^2y)/(x^3-1-2sqrt(x^3y-y))# Differentiating with respect to #x#: #-y'=1-1/2(x^3y-y)^{-1/2}(3x^2y+x^3y'-y')# Multiplying by #2sqrt(x^2y-y)# #x^3y'-y'-2y'sqrt(x^3y-y)=2sqrt(x^3y-y)-3x^2y# and this is #y'(x^3-1-2sqrt(x^3y-y))=2sqrt(x^3y-y)-3x^2y# so #y'=(2sqrt(x^3y-y)-3x^2y)/(x^3-1-2sqrt(x^3y-y))# if the denominator is not equal zero

Charles Arocha · Answer

undefined To implicitly differentiate the equation -y = x - sqrt(x^3y - y), we can follow these steps:

1. Differentiate each term of the equation with respect to $ x $.
2. Apply the chain rule whenever necessary.
3. Solve for $ \frac{{dy}}{{dx}} $.

Differentiating each term:

- Differentiating $ -y $ with respect to $ x $ gives $ -\frac{{dy}}{{dx}} $.
- Differentiating $ x $ with respect to $ x $ gives $ 1 $.
- Differentiating $ \sqrt{x^3y - y} $ with respect to $ x $ requires the chain rule.

Differentiating the square root term $ \sqrt{u} $ with respect to $ x $ (where $ u = x^3y - y $):

- The derivative of $ \sqrt{u} $ with respect to $ u $ is $ \frac{{1}}{{2\sqrt{u}}} $.
- Applying the chain rule, the derivative with respect to $ x $ is $ \frac{{1}}{{2\sqrt{u}}} \cdot \frac{{du}}{{dx}} $.

Now, we need to find $ \frac{{du}}{{dx}} $ using the product rule:

- Let $ u = x^3y - y $.
- Differentiating $ x^3y $ with respect to $ x $ gives $ 3x^2y + x^3\frac{{dy}}{{dx}} $.
- Differentiating $ -y $ with respect to $ x $ gives $ -\frac{{dy}}{{dx}} $.

Now, we can substitute the derivatives into the expression for $ \frac{{du}}{{dx}} $:

$$ \frac{{du}}{{dx}} = (3x^2y + x^3\frac{{dy}}{{dx}}) - \frac{{dy}}{{dx}} $$

Now, substitute all derivatives back into the original equation and solve for $ \frac{{dy}}{{dx}} $:

$$ -\frac{{dy}}{{dx}} = 1 - \frac{{1}}{{2\sqrt{x^3y - y}}} \left(3x^2y + x^3\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}}\right) $$

$$ -\frac{{dy}}{{dx}} = 1 - \frac{{3x^2y}}{{2\sqrt{x^3y - y}}} - \frac{{x^3}}{{2\sqrt{x^3y - y}}} \frac{{dy}}{{dx}} + \frac{{1}}{{2\sqrt{x^3y - y}}} \frac{{dy}}{{dx}} $$

$$ -\frac{{dy}}{{dx}} + \frac{{x^3}}{{2\sqrt{x^3y - y}}} \frac{{dy}}{{dx}} - \frac{{1}}{{2\sqrt{x^3y - y}}} \frac{{dy}}{{dx}} = 1 - \frac{{3x^2y}}{{2\sqrt{x^3y - y}}} $$

$$ \left(1 - \frac{{x^3}}{{2\sqrt{x^3y - y}}}\right) \frac{{dy}}{{dx}} = 1 - \frac{{3x^2y}}{{2\sqrt{x^3y - y}}} $$

$$ \frac{{dy}}{{dx}} = \frac{{1 - \frac{{3x^2y}}{{2\sqrt{x^3y - y}}}}}{{1 - \frac{{x^3}}{{2\sqrt{x^3y - y}}}}} $$