Question
Find the second derivative $y^{\prime \prime}(x)$.$$x^{2} y^{2}+3 x-4 y=5$$
Step 1
The derivative of $x^{2} y^{2}$ is $2x y^{2} + x^{2} 2y y'$, the derivative of $3x$ is $3$, and the derivative of $-4y$ is $-4y'$. Setting the derivative equal to zero, we get: $$2x y^{2} + x^{2} 2y y' + 3 - 4y' = 0$$ Show more…
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