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$\mathrm{m}$ is the slope of a tangent to the curve $\mathrm{e}^{\mathrm{y}}=1+\mathrm{x}^2$ at $\mathrm{x}=1$ then $\mathrm{m}=$
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Verified Answer
The correct answer is:
1
We are given that the curve is:
$$
\begin{aligned}
& \mathrm{e}^{\mathrm{y}}=1+\mathrm{x}^2 \\
& \Rightarrow \mathrm{y}=\log _{\mathrm{e}}\left|1+\mathrm{x}^2\right|
\end{aligned}
$$
Now, $\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{1}{1+\mathrm{x}^2}\right) \cdot 2 \mathrm{x}$
At $\mathrm{x}=1$
$$
\mathrm{m}=\left(\frac{1}{1+1}\right) \cdot 2=\frac{1}{2} \times 2 \ldots
$$
$$
\begin{aligned}
& \mathrm{e}^{\mathrm{y}}=1+\mathrm{x}^2 \\
& \Rightarrow \mathrm{y}=\log _{\mathrm{e}}\left|1+\mathrm{x}^2\right|
\end{aligned}
$$
Now, $\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{1}{1+\mathrm{x}^2}\right) \cdot 2 \mathrm{x}$
At $\mathrm{x}=1$
$$
\mathrm{m}=\left(\frac{1}{1+1}\right) \cdot 2=\frac{1}{2} \times 2 \ldots
$$
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