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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives3 Marks
Question
Differentiate the following from first principle:$\frac{\text{2x}+3}{\text{x}-2}$
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Answer

We have, $\text{f(x)}=\frac{2\text{x}+3}{\text{x}-2}$
$\therefore\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(a+h)}-\text{f(a)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(\frac{2\text{x}+2\text{h}+3}{\text{x+h}-2}\Big)-\Big(\frac{2\text{x}+3}{\text{x}-2}\Big)}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}^2+2\text{xh}+3\text{x}-4\text{x}-4\text{h}-6-2\text{x}^2-2\text{hx}+4\text{x}-3\text{x}-3\text{h}+6}{\text{h}(\text{x+h}-2)(\text{x}-2)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-7}{(\text{x+h}-2)(\text{x}-2)}$
$=\frac{-7}{(\text{x}-2)^2}$

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