CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$ y = xn + nx + xx + nn
✓
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, $\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{x})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{n}^\text{n})$ $=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{e}^{\log\text{x}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{x}+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big]$ $=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\Big[\text{x}\big(\frac{1}{\text{x}}\big)+\log\text{x}\Big]$ $=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[1+\log\text{x}\big]$ $=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\big[\log\text{e}+\log\text{x}\big] \\ \big[\because\log_\text{e}\text{e}=1\text{ and }\log\text{A}+\log\text{B}=\log(\text{AB})\big]$ $=\text{nx}^{\text{n}-1}+\text{n}^\text{x}\log\text{n}=\text{x}^{\text{x}}\log\big(\text{ex}\big)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.