If f(x)=3^x and g(x)=4^x,then frac{f^{prime}( | vedclass

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(A) Given $f(x) = 3^x$,the derivative is $f^{\prime}(x) = 3^x \log 3$. Evaluating at $x = 0$,we get $f^{\prime}(0) = 3^0 \log 3 = \log 3$.Given $g(x)

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$\frac{\log \left(\frac{3}{4}\right)}{1+(\log 3)(\log 4)}$

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(A) Given $f(x) = 3^x$,the derivative is $f^{\prime}(x) = 3^x \log 3$. Evaluating at $x = 0$,we get $f^{\prime}(0) = 3^0 \log 3 = \log 3$.Given $g(x) = 4^x$,the derivative is $g^{\prime}(x) = 4^x \log 4$. Evaluating at $x = 0$,we get $g^{\prime}(0) = 4^0 \log 4 = \log 4$.Substituting these values into the expression,we have $\frac{f^{\prime}(0)-g^{\prime}(0)}{1+f^{\prime}(0) g^{\prime}(0)} = \frac{\log 3 - \log 4}{1 + (\log 3)(\log 4)}$.Using the logarithmic property $\log a - \log b = \log \left(\frac{a}{b}\right)$,the numerator becomes $\log \left(\frac{3}{4}\right)$.Thus,the expression is $\frac{\log \left(\frac{3}{4}\right)}{1 + (\log 3)(\log 4)}$.

If $f(x)=3^x$ and $g(x)=4^x$,then $\frac{f^{\prime}(0)-g^{\prime}(0)}{1+f^{\prime}(0) g^{\prime}(0)}$ is

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