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Derivative from First Principle

Find the deri...

Question

Find the derivative by the first principle of $f (x) = x^{3} + 4 x$

3 x^{2} + 4

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x \sqrt{x}

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\frac{1}{2 x + 3}

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\frac{x - 1}{2 x + 7}

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Solution

The correct option is A $3 x^{2} + 4$
$f (x) = x^{3} + 4 x$

$f^{'} (x) = l i m_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= l i m_{h \to 0} \frac{(x + h)^{3} + 4 (x + h) - x^{3} - 4 x}{h} = l i m_{h \to 0} \frac{x^{3} + 3 x^{2} h + 3 x h^{2} + h^{3} + 4 x + 4 h - x^{3} - 4 x}{h}$

$= l i m_{h \to 0} \frac{(x + h)^{3} + 4 (x + h) - x^{3} - 4 x}{h} = l i m_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3} + 4 h}{h} = l i m_{h \to 0} \frac{h (3 x^{2} + 4 + h^{2} + 3 x h)}{h} = 3 x^{2} + 4$

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