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If $f(x)=\left(x-x_{0}\right) \phi(x)$ and $\phi(x)$ is continuous at $x=x_{0}$, then what is $\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)$ equal to ?
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The correct answer is:
$\phi\left(\mathrm{x}_{0}\right)$
$\quad$ Given that $\mathrm{f}(\mathrm{x})=\left(\mathrm{x}-\mathrm{x}_{0}\right) \phi(\mathrm{x})$
Differentiating w. r. t. $\mathrm{x}$ $\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{x}-\mathrm{x}_{0}\right) \phi^{\prime}(\mathrm{x})+\phi(\mathrm{x})(1)$
Putting $\mathrm{x}=\mathrm{x}_{0}$
$\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)=\phi\left(\mathrm{x}_{0}\right)$
Differentiating w. r. t. $\mathrm{x}$ $\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{x}-\mathrm{x}_{0}\right) \phi^{\prime}(\mathrm{x})+\phi(\mathrm{x})(1)$
Putting $\mathrm{x}=\mathrm{x}_{0}$
$\mathrm{f}^{\prime}\left(\mathrm{x}_{0}\right)=\phi\left(\mathrm{x}_{0}\right)$
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