Exercise
Given the following function
y=\frac{1}{x^2-1}
Prove that the following holds:
2y'^2-y\cdot y''=2y^3
Proof
First, we compute the first derivative and the second derivative, since they appear in the equation that needs to be proved.
y=\frac{1}{x^2-1}
Using Derivative formulas and the quotient rule in Derivative Rules, we get the derivative:
y'=\frac{-2x}{{(x^2-1)}^2}
We want to compute the second derivative. To do this, we derive the first derivative and get:
y''=\frac{-2{(x^2-1)}^2+2x\cdot 2(x^2-1)\cdot 2x}{{(x^2-1)}^4}=
We simplify the second derivative:
=\frac{-2{(x^2-1)}^2+8x^2\cdot (x^2-1)}{{(x^2-1)}^4}=
=\frac{(x^2-1)(-2(x^2-1)+8x^2)}{{(x^2-1)}^4}=
=\frac{-2x^2+2+8x^2}{{(x^2-1)}^3}=
=\frac{6x^2+2}{{(x^2-1)}^3}
We set the function and the derivative on the left side of the equation we need to prove, and we want to get the expression 0n the right side .
2y'^2-y\cdot y''=
=2\cdot {(\frac{-2x}{{(x^2-1)}^2})}^2-\frac{1}{x^2-1}\cdot \frac{6x^2+2}{{(x^2-1)}^3}=
=\frac{8x^2}{{(x^2-1)}^4}-\frac{6x^2+2}{{(x^2-1)}^4}=
=\frac{8x^2-6x^2-2}{{(x^2-1)}^4}=
=\frac{2x^2-2}{{(x^2-1)}^4}=
=\frac{2(x^2-1)}{{(x^2-1)}^4}=
=\frac{2}{{(x^2-1)}^3}=
=2y^3
We were able to reach the right side of the equation.
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