LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
Find the derivative of the following function from first principle. $(\text{x}-1)(\text{x}-2)$
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Answer
Let $\text{f}(\text{x})=(\text{x}-1)(\text{x}-2)$. Accordingly,from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}+\text{h}-1)(\text{x}+\text{h}-2)-(\text{x-1})(\text{x}-2)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}^2+\text{h}\text{x}-2\text{x}+\text{h}\text{x}+\text{h}^2-2\text{h}-\text{x}-\text{h}+2)-(\text{x}^2-2\text{x}-\text{x}+2)}{\text{h}}$$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{h}\text{x}+\text{h}\text{x}+\text{h}^2-2\text{h}-\text{h})}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{(2\text{h}\text{x}+\text{h}^2-3\text{h})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}{(2\text{x+}\text{h}-3)}$$=(2\text{x}+0-3)$
$=2\text{x}-3$
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