6x²+3x5y²=0
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[tex] \frac{d}{dx} (6 \times ^{2}) + \frac{d}{dx} (3x {5y}^{2} ) = \frac{d}{dx} (0)[/tex]
Use differentuational rule
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[tex]( \frac{d}{dx} (a \times f) = a \times \frac{d}{dx} (f) \: )[/tex]
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[tex] \red{\frac{d}{dx} (6 \times ^{2})} + \frac{d}{dx} + (3x {5y}^{2} ) = \frac{d}{dx} (0)[/tex]
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[tex] \red{6 \times \frac{d}{dx}( {x}^{2}) } + \frac{d}{dx} (3x {5y}^{2} ) = \frac{d}{dx} (0)[/tex]
Use the chain rule
[tex] \frac{d}{dx} (3x {5y}^{2} ) = \frac{d}{dy} (3 \times {5y}^{2} ) \times \frac{dy}{dx} [/tex]
to take the deravative
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[tex] \frac{d}{dx} (6 \times ^{2}) \red{+ \frac{d}{dx} (3x {5y}^{2} )} = \frac{d}{dx} (0)[/tex]
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[tex] 6 \times \frac{d}{dx}( {x}^{2}) \red{ + \frac{d}{dy} (3x {5y}^{2} ) \times \frac{dy}{dx} } = \frac{d}{dx} (0)[/tex]
The deravative of a constant is always zero
[tex] \frac{d}{dx} (6 \times ^{2}) \frac{d}{dx} (3x {5y}^{2} ) = \red{ \frac{d}{dx} (0)}[/tex]
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[tex] 6 \times \frac{d}{dx}( {x}^{2}) + \frac{d}{dy} (3x {5y}^{2} ) \times \frac{dy}{dx} = \frac{d}{dx} \red{ = 0}[/tex]
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[tex]\huge\blue{\boxed{\tt{\colorbox{pink}{Final Answer}}}}[/tex]
[tex]\red{\boxed{\tt{\colorbox{pink}{↓}}}}[/tex]
[tex] \frac{dy}{dc} = - \frac{2x}{5y} [/tex]
