How do you find #(d^2y)/(dx^2)# for #-2y^2+2=3x#?
Please see the explanation section below.
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To find (\frac{{d^2y}}{{dx^2}}) for (-2y^2+2=3x), we first differentiate the given equation with respect to (x) twice, using implicit differentiation. This involves finding (\frac{{dy}}{{dx}}) and then differentiating it again to find (\frac{{d^2y}}{{dx^2}}).
The first derivative (\frac{{dy}}{{dx}}) is found by differentiating both sides of the equation with respect to (x): [ \frac{{d(-2y^2+2)}}{{dx}} = \frac{{d(3x)}}{{dx}} ] This simplifies to: [ -4y\frac{{dy}}{{dx}} = 3 ] Solving for (\frac{{dy}}{{dx}}), we get: [ \frac{{dy}}{{dx}} = \frac{{3}}{{-4y}} ]
Now, we differentiate (\frac{{dy}}{{dx}}) with respect to (x) again to find (\frac{{d^2y}}{{dx^2}}): [ \frac{{d\left(\frac{{dy}}{{dx}}\right)}}{{dx}} = \frac{{d\left(\frac{{3}}{{-4y}}\right)}}{{dx}} ] After simplifying and substituting (\frac{{dy}}{{dx}}), we get: [ \frac{{d^2y}}{{dx^2}} = \frac{{12}}{{4y^2}} = \frac{{3}}{{y^2}} ]
Therefore, (\frac{{d^2y}}{{dx^2}} = \frac{{3}}{{y^2}}) for the given equation (-2y^2+2=3x).
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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.
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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