Calculating Derivative – Proof of an equation with derivatives – Exercise 6284

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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