Differentiate from first principle:

$(x i i i) (x^{2} + 1) (x - 5)$

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Solution

Given:

$f (x) = (x^{2} + 1) (x - 5)$

$= x^{3} - 5 x^{2} + x - 5$

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:

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(x + h)^{3} - 5 (x + h)^{2} + (x + h) - 5 - (x^{3} - 5 x^{2} + x - 5)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(x + h)^{3} - x^{3} - 5 {(x + h)^{2} - x^{2}} + h}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{(h) {(x + h)^{2} + x^{2} + x (x + h)} - 5 (h) (2 x + h) + h}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 [(x + h)^{2} + x^{2} + x (x + h) - 5 (2 x + h) + 1]$

$\Rightarrow f^{'} (x) = x^{2} + x^{2} + x^{2} - 5 (2 x) + 1$

$\Rightarrow f^{'} (x) = 3 x^{2} - 10 x + 1$

Hence, the derivative of $(x^{2} + 1) (x - 5)$ is $3 x^{2} - 10 x + 1$

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