Differentiate from first principle:

$(x i i) x^{3} + 4 x^{2} + 3 x + 2$

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Solution

Givne:

$f (x) = x^{3} + 4 x^{2} + 3 x + 2$

The derivative of a function $f (x)$ is defined as:

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

Putting $f (x)$ in the above expression, we get:

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(x + h)^{3} + 4 (x + h)^{2} + 3 (x + h) + 2 - (x^{3} + 4 x^{2} + 3 x + 2)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{[(x + h)^{3} - x^{3}] + 4 [(x + h)^{2} - x^{2}] + 3 h}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{[h {(x + h)^{2} + x^{2} + x (x + h)}] 4 h (2 x + h) + 3 h}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 [(x + h)^{2} + x^{2} + x (x + h) + 4 (2 x + h) + 3]$

$\Rightarrow f^{'} (x) = x^{2} + x^{2} + x^{2} + 8 x + 3$

$\Rightarrow f^{'} (x) = 3 x^{2} + 8 x + 3$

Hence, the derivative of $x^{3} + 4 x^{2} + 3 x + 2$ is $3 x^{2} + 8 x + 3$

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