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

Taken you to a point where you can take over. Hopefully my workings are correct.

Consider:

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Second differential

So for the second differential we have:

I will let you take it from this point

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To find ( \frac{{d^2y}}{{dx^2}} ) for ( x^3 = 2y^2 + 5 ), first, differentiate the given equation implicitly with respect to ( x ) twice, then solve for ( \frac{{d^2y}}{{dx^2}} ).

- Differentiate the given equation implicitly with respect to ( x ) once:

[ \frac{{d}}{{dx}}(x^3) = \frac{{d}}{{dx}}(2y^2 + 5) ]

[ 3x^2 = 4y\frac{{dy}}{{dx}} ]

- Now, differentiate the obtained equation with respect to ( x ) again:

[ \frac{{d}}{{dx}}(3x^2) = \frac{{d}}{{dx}}(4y\frac{{dy}}{{dx}}) ]

[ 6x = 4\left(\frac{{dy}}{{dx}}\right)^2 + 4y\frac{{d^2y}}{{dx^2}} ]

- Now, solve for ( \frac{{d^2y}}{{dx^2}} ):

[ 6x = 4\left(\frac{{dy}}{{dx}}\right)^2 + 4y\frac{{d^2y}}{{dx^2}} ]

[ 4y\frac{{d^2y}}{{dx^2}} = 6x - 4\left(\frac{{dy}}{{dx}}\right)^2 ]

[ \frac{{d^2y}}{{dx^2}} = \frac{{6x - 4\left(\frac{{dy}}{{dx}}\right)^2}}{{4y}} ]

This gives you the second derivative ( \frac{{d^2y}}{{dx^2}} ) of the implicit function ( y ) with respect to ( x ).

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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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