Let:
x = 4 $\text{x}+\triangle \text{x}=4.04$Then,
$\triangle\text{x} =0.04$ For x = 4 $\text{y}=\log_\text {e}4=\frac{\log_{10}4}{\log_{10}\text {e}}=\frac{0.6021}{0.4343}=1.386368$Let:
$\text{dx}=\triangle \text{x}=0.04$Now, $\text{y}=\log_\text {e}\text{x}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{\text {x}}$ $\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x}=4} =\frac{1}{4}$ $\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {4}\times0.04=0.01$ $\Rightarrow\triangle \text{y}=0.01$ $\therefore\log_\text {e}4.04=\text{y}+\triangle\text{y} =1.39638$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\int\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\text{dx}$