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Calculus help because I am stuck on this question

Calculus help because I am stuck on this question-example-1
User MaxNoe
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1 Answer

5 votes


y=\left((x^2+2)/(x^2-2)\right)^6

Chain and power rule:

Let
u=(x^2+2)/(x^2-2). By the chain rule,


y=u^6\implies y'=6u^5u'

Quotient rule:


u=(x^2+2)/(x^2-2)\implies u'=((x^2-2)(x^2+2)'-(x^2+2)(x^2-2)')/((x^2-2)^2)


u'=(2x(x^2-2)-2x(x^2+2))/((x^2-2)^2)=-(8x)/((x^2-2)^2)

Putting everything together, we have


y'=6\left((x^2+2)/(x^2-2)\right)^5\left(-(8x)/((x^2-2)^2)\right)


y'=-(48x(x^2+2)^5)/((x^2-2)^7)

User Havij
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