What is the implicit derivative of #3=1/y -x^2 #?

Question

Charles Arocha · Accepted Answer

#dy/dx = -2xy^2# The trick is in finding the derivative of each term, for term containing #y# find derivative as usual and put #dy/dx# next to it to indicate it is differentiated with respect to #x#

#3 = 1/y - x^2# #3 = y^-1 - x^2# Differentiate both sides with respect to #x# #d/dx(3) = d/dx(y^-1) - d/dx(x^2)# #0 = -y^(-1-1)dy/dx - 2x# #0=-y^-2dy/dx -2x# #0= -1/y^2 dy/dx -2x# Add 2x to both the sides #2x = -1/y^2 dy/dx# Multiply both sides by #-y^2# #-2xy^2 = dy/dx# Answer #dy/dx = -2xy^2#

Isaiah Allen · Answer

undefined To find the implicit derivative of $3 = \frac{1}{y} - x^2$:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the chain rule and product rule where necessary.
3. Solve for $\frac{dy}{dx}$ to find the implicit derivative.

Differentiating $3 = \frac{1}{y} - x^2$ with respect to $x$:

$$0 = -\frac{1}{y^2} \frac{dy}{dx} - 2x$$

Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = -\frac{2xy^2}{y^2} = -\frac{2x}{y^2}$$

Luke Alvarez · Answer

undefined To find the implicit derivative of $3 = \frac{1}{y} - x^2$, differentiate both sides of the equation with respect to $x$. Then, solve for $\frac{dy}{dx}$, the derivative of $y$ with respect to $x$.

Differentiating both sides with respect to $x$, we get:

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

Solving this, we find:

$$0 = -\frac{1}{y^2}\frac{dy}{dx} - 2x$$

Rearranging and solving for $\frac{dy}{dx}$, we get:

$$\frac{dy}{dx} = -\frac{2xy^2}{1}$$