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Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSLimits and Derivatives5 Marks
Question
Differentiate the functions with respect to 'x'.
$\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}$
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Answer

Let $\text{f}(\text{x})=\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}\ ...(\text{i})$
$\Rightarrow\text{f}(\text{x}+\Delta\text{x})=\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}\ ...(\text{ii})$
Subtracting eq. (i) from eq. (ii)
$\Rightarrow\text{f}(\text{x}+\Delta\text{x}-\text{f}(\text{x})=\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}-\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}$
Dividing both sides by taken the limit, we get
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\text{f}(\text{x}+\Delta\text{x})-\text{f}(\text{x})}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{\frac{\text{a}(\text{x}+\Delta\text{x})+\text{b}}{\text{c}(\text{x}+\Delta\text{x})+\text{d}}-\frac{\text{ax}+\text{b}}{\text{cx}+\text{d}}}{\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{-\text{acx}^{2}-\text{ac}\Delta\text{x}\cdot-\text{adx}-\text{bcx}-\text{bc}\cdot\Delta\text{x}-\text{bd}}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d}).\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{(\text{ad}-\text{bc})\Delta\text{x}}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d})\cdot.\Delta\text{x}}$
$=\lim\limits_{\Delta\text{x} \rightarrow 0}\frac{(\text{ad}-\text{bc})}{(\text{cx}+\text{c}\Delta\text{x}+\text{d})(\text{cx}+\text{d})}$
Taking limit, we get
$=\frac{(\text{ad}-\text{bc})}{(\text{cx}+\text{d})(\text{cx}+\text{d})}=\frac{\text{ad}-\text{bc}}{(\text{cx}+\text{d})^{2}}$
Hence, the required answer is $\frac{\text{ad}}{(\text{cx}+\text{d})^{2}}.$

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