If $\mathrm{f}(x)=3^x ; \mathrm{g}(x)=4^x$, then…

If $\mathrm{f}(x)=3^x ; \mathrm{g}(x)=4^x$, then $\frac{\mathrm{f}^{\prime}(0)-\mathrm{g}^{\prime}(0)}{1+\mathrm{f}^{\prime}(0) \mathrm{g}^{\prime}(0)}$ is
  1. $\frac{\log \left(\frac{3}{4}\right)}{1+(\log 3)(\log 4)}$
  2. $\frac{\log \left(\frac{3}{4}\right)}{1+\log 12}$
  3. $\frac{\log 12}{1+\log 12}$
  4. $\frac{\log \left(\frac{3}{4}\right)}{1-\log 12}$

Solution

$\begin{aligned} \mathrm{f}^{\prime}(x)=3^x \log 3 & \Rightarrow \mathrm{f}^{\prime}(0)=\log 3 \\ \mathrm{~g}^{\prime}(x)=4^x \log 4 & \Rightarrow \mathrm{g}^{\prime}(0)=\log 4 \\ \therefore \quad \frac{\mathrm{f}^{\prime}(0)-\mathrm{g}^{\prime}(0)}{1+\mathrm{f}^{\prime}(0) \mathrm{g}^{\prime}(0)} & =\frac{\log 3-\log 4}{1+(\log 3)(\log 4)} \\ & =\frac{\log \left(\frac{3}{4}\right)}{1+(\log 3)(\log 4)}\end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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