If $f(x)=\log \left\{\frac{u(x)}{v(x)}\right\}, u(1)=v(1)$ and $u^{\prime}(1)=v^{\prime}(1)=2$, then find the value of $f^{\prime}(1)$.
Using the Chain Rule of Differentiation,
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\frac{\mathrm{u}(\mathrm{x})}{\mathrm{v}(\mathrm{x})}} \cdot \frac{\mathrm{v}(\mathrm{x}) \cdot \mathrm{u}^{\prime}(\mathrm{x})-\mathrm{v}^{\prime}(\mathrm{x}) \cdot \mathrm{u}(\mathrm{x})}{(\mathrm{v}(\mathrm{x}))^{2}}$
$=\frac{v(x) \cdot u^{\prime}(x)-v^{\prime}(x) \cdot u(x)}{u(x) \cdot v(x)}$
Putting $x=1$,
$\mathrm{f}^{\prime}(1)=\frac{\mathrm{v}(1) \cdot \mathrm{u}^{\prime}(1)-\mathrm{v}^{\prime}(1) \cdot \mathrm{u}(1)}{\mathrm{u}(1) \cdot \mathrm{v}(1)}$
$=\frac{2 \mathrm{v}(1)-2 \mathrm{u}(1)}{\mathrm{u}(1) \cdot \mathrm{v}(1)}$
Since, $u(1)=v(1)$,
$2 v(1)-2 u(1)=0$
i.e., $f^{\prime}(1)=0$ (Ans)
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