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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives2 Marks
Question
Find the derivative of (x -1) (x - 2) from first principle.
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Answer

We have, f(x) = (x - 1)(x - 2)
= x2 - 3x + 2
By first principle of derivative, we have
${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {{{(x + h)}^2} - 3(x + h) + 2} \right] - \left[ {{x^2} - 3x + 2} \right]}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\left( {{x^2} + {h^2} + 2xh - 3x - 3h + 2} \right] - \left[ {{x^2} - 3x + 2} \right]} \right.}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2hx + {h^2} - 3h}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{h(2x + h - 3)}}{h}$
$= 2 x - 3$

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