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Find y': 2y - 5x^3 y^2 + 1 = 0 .

Question:

Find {eq}y': 2y - 5x^3 y^2 + 1 = 0 {/eq}.

Derivative of Function:

The derivative of function f(x) at point 'a' is

{eq}f'(a) =\lim _{h\rightarrow0}\frac{f(a+h) - f(a) }{h} {/eq}

provided this limit exists. If this limit exists then we say f is differentiable at point x=a.

Answer and Explanation:

{eq}2y - 5x^3 y^2 + 1 = 0 {/eq}

Differentiate with respect to x , we get

{eq}\frac{d}{dx}(2y - 5x^3 y^2 + 1 )=0\\ 2y' - 15x^2 y^2=0 {/eq}

{eq}y' = \frac{dy}{dx} {/eq}

Therefore {eq}y'= \frac{15x^2 y^2}{2} {/eq}


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