# How do you find #(d^2y)/(dx^2)# for #5x+3y^2=1#?

Use product rule, chain rule and sum rule.

Differentiating again using the product rule (chain rule is applied in the first term, hence the squared derivative):

Rearranging the equation for the first derivative:

Substituting this into the equation for the second derivative:

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To find (\frac{d^2y}{dx^2}) for the equation (5x + 3y^2 = 1), we first need to find (\frac{dy}{dx}) by implicitly differentiating the equation with respect to (x). Then, we differentiate (\frac{dy}{dx}) with respect to (x) again to find (\frac{d^2y}{dx^2}).

First, differentiate the equation (5x + 3y^2 = 1) with respect to (x): [5 + 6y\frac{dy}{dx} = 0]

Solve for (\frac{dy}{dx}): [\frac{dy}{dx} = \frac{-5}{6y}]

Now, differentiate (\frac{dy}{dx}) with respect to (x) to find (\frac{d^2y}{dx^2}): [\frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dx} \left(\frac{-5}{6y}\right)] [\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{-5}{6y}\right)] [\frac{d^2y}{dx^2} = \frac{0 - (-5)\left(\frac{d}{dx}(6y)\right)}{(6y)^2}] [\frac{d^2y}{dx^2} = \frac{5 \cdot 6\frac{dy}{dx}}{(6y)^2}]

Substitute the expression for (\frac{dy}{dx}) into the equation: [\frac{d^2y}{dx^2} = \frac{5 \cdot 6 \cdot \left(\frac{-5}{6y}\right)}{(6y)^2}]

Simplify: [\frac{d^2y}{dx^2} = \frac{-25}{(6y)^3}]

Thus, (\frac{d^2y}{dx^2} = \frac{-25}{216y^3}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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