0Derivative of x³ from first principle
If y = f(x) => x³
dy/dx = Lt [f(x+∆x) - f(x)]/∆x
∆x => 0
here f(x) = x³
Therefore y' = Lt [ f(x + ∆x)³ - x³ ] / ∆x
∆x=> 0
=> Lt [ x³ + ∆x³ + 3x∆x(x + ∆x) - x³ ]/∆x
∆x => 0
=> Lt [ x³ + ∆x³ + 3x²∆x + 3x∆x² - x³ ] / ∆x
∆x => 0
=> Lt [ ∆x³ + 3x²∆x + 3x∆x² ] / ∆x
∆x => 0
=> Lt [ ∆x³/∆x + 3x²∆x/∆x + 3x∆x²/∆x ]
∆x => 0
=> Lt [∆x³/∆x] + Lt [3x²∆x/∆x] + Lt [3x∆x²/∆x]
∆x=>0 ∆x=> 0 ∆x=> 0
=> Lt ∆x² + Lt 3x² + Lt 3x∆x
∆x=>0 ∆x=>0 ∆x=>0
=> 0 + 3x² + 0
=> 3x²
