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Derivatives — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives4 Marks
Question
Differentiate the following from first principle$\text{a}^{\sqrt{\text{x}}}$
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Answer

$\text{f}(\text{x})=\text{a}^\sqrt{\text{x}}=\text{e}^{\sqrt{\text{x}}\log\text{a}}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{\text{x}+\text{h}}\log\text{a}}-\text{e}^{\sqrt{\text{x}}\log\text{a}-1}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{\sqrt{\text{x}+\text{h}}\log\text{a}-\sqrt{\text{x}}\log\text{a}}-1}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{\text{h}}$
Multiply numerator and denominator by $(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}$
$\text{f}(\text{x})=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}-1}{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}}$
Multiply numerator and denominator by $(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})$
$\text{f}(\text{x})=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\lim_\limits{\text{h}\rightarrow0}\log\text{a}\frac{\text{h}}{\text{h}(\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}})}$
$=\text{e}^{\sqrt{\text{x}}\log\text{a}}\frac{\log\text{a}}{2\sqrt{\text{x}}}$
$=\frac{\text{a}^{\sqrt{\text{x}}}}{2\sqrt{\text{x}}}\log_\text{e}\text{a}$

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