frac{d}{dx}(log_5 x^2) = . . . . . . | vedclass

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(B) We know that $\log_a b = \frac{\log_e b}{\log_e a}$.So,$\log_5 x^2 = \frac{\ln x^2}{\ln 5} = \frac{2 \ln x}{\ln 5}$.Now,differentiating with respe

Vedclass · Accepted Answer

$\frac{2}{(\log 5)x}$

Vedclass · Answer

(B) We know that $\log_a b = \frac{\log_e b}{\log_e a}$.So,$\log_5 x^2 = \frac{\ln x^2}{\ln 5} = \frac{2 \ln x}{\ln 5}$.Now,differentiating with respect to $x$:$\frac{d}{dx} \left( \frac{2 \ln x}{\ln 5} \right) = \frac{2}{\ln 5} \cdot \frac{d}{dx}(\ln x)$.Since $\frac{d}{dx}(\ln x) = \frac{1}{x}$,we get:$\frac{2}{\ln 5} \cdot \frac{1}{x} = \frac{2}{x \ln 5}$.Since $\ln 5 = \log 5$,the result is $\frac{2}{x \log 5}$.

$\frac{d}{dx}(\log_5 x^2) = $ . . . . . .

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