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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
y = xn + nx + xx + nn
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Answer

We have, y = xn + nx + xx + nx
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\log\text{x}^\text{x}}+\text{n}^\text{n}$
$\Rightarrow\text{y}=\text{x}^\text{n}+\text{n}^\text{x}+\text{e}^{\text{x}\log\text{x}}+\text{n}^\text{n}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{n})$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{e}^{\log\text{x}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\Big[\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}\Big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[1+\log\text{x}\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[\log\text{e}+\log\text{x}\big] \\ \big[\because\log_\text{e}\text{e}=1\text{ and }\log\text{A}+\log\text{B}=\log(\text{AB})\big]$
$=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\log\big(\text{ex}\big)$

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