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0 Finding derivative of x² from the first principles
f(x) = x²
Let y = x²
From the first principle
dy/dx = Lt [f(x+∆x) - f(x)]/∆x ]
∆x=>0
=> Lt ∆x=> 0 [(x + ∆x)² - x²]/∆x
=> Lt ∆x=> 0 [x² + 2x∆x + ∆x² - x²]/∆x
=> Lt ∆x=>0 [2x∆x + ∆x²]/∆x
=> Lt ∆x => 0 [(2x∆x/∆x) + (∆x²/∆x)]
=> Lt ∆x=> 0 [2x + ∆x]
=> 2x + 0
=> 2x
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