# Find second derivative of #y# if #x^6+y^6=1#?

To find the derivative we will use implicit differentiation.

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To find the second derivative of ( y ) with respect to ( x ) given the equation ( x^6 + y^6 = 1 ), we'll differentiate implicitly twice.

First, differentiate the equation with respect to ( x ) to find ( \frac{{dy}}{{dx}} ). Then, differentiate ( \frac{{dy}}{{dx}} ) again with respect to ( x ) to find ( \frac{{d^2y}}{{dx^2}} ).

The given equation is ( x^6 + y^6 = 1 ).

Differentiating once with respect to ( x ): [ 6x^5 + 6y^5 \frac{{dy}}{{dx}} = 0 ] [ \frac{{dy}}{{dx}} = -\frac{{x^5}}{{y^5}} ]

Now, differentiate again with respect to ( x ): [ \frac{{d^2y}}{{dx^2}} = -\frac{{5x^4}}{{y^5}} - \frac{{6y^5}}{{x^4}} \left( \frac{{dy}}{{dx}} \right)^2 ]

Substitute ( \frac{{dy}}{{dx}} = -\frac{{x^5}}{{y^5}} ) into the equation: [ \frac{{d^2y}}{{dx^2}} = -\frac{{5x^4}}{{y^5}} - \frac{{6y^5}}{{x^4}} \left( -\frac{{x^5}}{{y^5}} \right)^2 ]

Simplify: [ \frac{{d^2y}}{{dx^2}} = -\frac{{5x^4}}{{y^5}} - \frac{{6x^{10}}}{{y^{15}}} ]

This is the second derivative of ( 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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