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

1 Answer

Jun 10, 2018

#\frac{dy}{dx}=-\frac{3x^2}{4y}#

Explanation:

Differentiate both sides as such:
#\frac{d}{dx}(-2y^{2}+3)=\frac{d}{dx}(x^{3})#
this leads to:

#-2\frac{d}{dx}(y^{2})=3x^{2}#

#-2(2y\frac{dy}{dx})=3x^{2}#

rearrange to make #\frac{dy}{dx}# the factor to get:

#\frac{dy}{dx}=-\frac{3x^{2}}{4y}#

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