Find the derivative of x+1/x-1 using first principle
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Step-by-step explanation:
We have:
f(x)=x−1x+1⇒f(x+h)=x+h−1x+h+1
By first principle, we get:
f'(x)=limh→0 f(x+h)−f(x)h=x+h−1x+h+1−x−1x+1h=limh→0(x+h−1)(x+1)−(x+h+1)(x−1)(x+h+1)(x+1)h=limh→0x2+x+hx+h−x−1−(x2−x+hx−h+x−1)(x+h+1)(x+1)hlimh→0x2+x+hx+h−x−1−x2+x−hx+h−x+1(x+h+1)(x+1)h=limh→0h+h(x+h+1)(x+1)h=limh→02h(x+h+1)(x+1)h=limh→02(x+h+1)(x+1)
Apply limitf'(x)=2(x+0+1)(x+1)⇒f'(x)=2(x+1)2
it's hard to write in this
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