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Derivative using first principles

P
Pramod .
24/09/2021 0 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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P

Pramod .

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Derivative using first principles
Derivative of x³ from first principle If y = f(x) => x³ dy/dx = Lt /∆x ∆x => 0 here f(x) = x³ Therefore y' = Lt / ∆x ∆x=> 0 =>...
P

Pramod .

1 0
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