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Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsDerivatives2 Marks
Question
Differentiate the following function with respect to $(\text{x})$:$\text{e}^{\text{x}\log\text{a}}+\text{e}^{\text{a}\log\text{x}}+{\text{e}^{\text{a}\log\text{a}}}$
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Answer

We have,$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}}+\text{e}^{\text{a}\log\text{x}}+{\text{e}^{\text{a}\log\text{a}}})$
$=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{a}})$
$=\text{e}^{\text{x}\log\text{a}}.\log\text{a}+\text{e}^{\text{a}\log\text{x}}.\frac{\text{a}}{\text{x}}+0\ [\because\text{e}^{\text{a}\log\text{a}}$is constant$]$
$=\log\text{a}.\text{e}^{\text{x}\log\text{a}}+\frac{\text{a}}{\text{x}}\text{e}^{\text{a}\log\text{x}}$
$=\log\text{a}.\text{a}^\text{x}+\frac{\text{a}}{\text{x}}\text{x}^\text{a}\ [\text{a}^\text{x}$can be written as $\text{e}^{\text{x}\log\text{a}}]$
$=\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$

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