$\frac{d}{dx}(e^{x²} + x^{x²})$

$x(12-2x)²$ এর বৃহত্তম ও ক্ষুদ্রতম মান নির্ণয় করো

$\mathrm{y}=\mathrm{e}^{\mathrm{x}^{2}}+\mathrm{e}^{\ln \mathrm{x}^{\mathrm{x}^{2}}}=\mathrm{e}^{\mathrm{x}^{2}}+\mathrm{e}^{\mathrm{x}^{2} \ln \mathrm{x}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}^{2}} \cdot 2 \mathrm{x}+\mathrm{e}^{\mathrm{x}^{2} \ln \mathrm{x}}\left(\mathrm{x}^{2} \cdot \frac{1}{\mathrm{x}}+2 \mathrm{x} \cdot \ln \mathrm{x}\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{xe}^{\mathrm{x}^{2}}+\mathrm{x}^{\mathrm{x}^{2}}(\mathrm{x}+2 \mathrm{x} \ln \mathrm{x})$ Ans.

$\mathrm{x}(12-2 \mathrm{x})^{2}$ এর বৃহত্তম ও ক্ষুদ্রতম মান নির্ণয় কর।

Let, $\mathrm{f}(\mathrm{x})=\mathrm{x}(12-2 \mathrm{x})^{2}=\mathrm{x}\left(4 \mathrm{x}^{2}-48 \mathrm{x}+144\right)=4 \mathrm{x}^{3}-48 \mathrm{x}^{2}+144 \mathrm{x}$ $f^{\prime}(x)=12 x^{2}-96 x+144 ; f^{\prime \prime}(x)=24 x-96$

for minimum and maximum value

$\mathrm{f}^{\prime}(\mathrm{x})=12 \mathrm{x}^{2}-96 \mathrm{x}+144=0 \quad \therefore \mathrm{x}=6,2$

$\mathrm{f}^{\prime \prime}(2)=-48<0$; maximum value will be obtained. $\quad \therefore$ maximum value $\mathrm{f}(2)=128$

$\mathrm{f}^{\prime \prime}(6)=48>0$; minimum value obtained.

$\therefore$ minimum value $\mathrm{f}(6)=0$