How do you find #(dy)/(dx)# given #-3x^2y^2-2y^3+5=5x^2#?

Question

Samantha Adkison · Accepted Answer

#(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)#

we want

#d/(dx)(-3x^2y^2)-d/(dx)(2y^3)+d/(dx)(5)=d/(dx)(5x^2)#

We will differentiate implicitly.

The first term will also need the product rule

#color(red)(d/(dx)(uv)=v(du)/(dx)+u(dv)/(dx))#

#y^2(-6x)+(-3x^2)2y(dy)/(dx)-6y^2(dy)/(dx)+0=10x#

#-6xy^2-6x^2y(dy)/(dx)-6y^2(dy)/(dx)=10x#

now rearrange for #(dy)/(dx)# and tidy up.

#(dy)/(dx)(-6x^2y-6y^2)=10x+6xy^2#

#(dy)/(dx)=(10x+6xy^2)/(-6y(x^2+y)#

#(dy)/(dx)=(-(5x+3xy^2))/(3y(x^2+y)#

Emma Aldridge · Answer

undefined To find $\frac{{dy}}{{dx}}$ given the equation $-3x^2y^2 - 2y^3 + 5 = 5x^2$, we'll differentiate both sides of the equation implicitly with respect to $x$, then solve for $\frac{{dy}}{{dx}}$.

Starting with the given equation:

$$ -3x^2y^2 - 2y^3 + 5 = 5x^2 $$

Differentiating both sides with respect to $x$:

$$ \frac{{d}}{{dx}}(-3x^2y^2) + \frac{{d}}{{dx}}(-2y^3) + \frac{{d}}{{dx}}(5) = \frac{{d}}{{dx}}(5x^2) $$

Using the chain rule and product rule where necessary, we find the derivatives:

$$ -6xy^2 \frac{{dy}}{{dx}} - 6x^2y \frac{{dy}}{{dx}} - 6y^2\frac{{dy}}{{dx}} - 6y^2 + 0 = 10x $$

Rearranging terms and factoring out $\frac{{dy}}{{dx}}$:

$$ (-6xy^2 - 6x^2y - 6y^2) \frac{{dy}}{{dx}} = 10x + 6y^2 $$

Finally, solving for $\frac{{dy}}{{dx}}$:

$$ \frac{{dy}}{{dx}} = \frac{{10x + 6y^2}}{{-6xy^2 - 6x^2y - 6y^2}} $$