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Question Number 169366 by Mastermind last updated on 29/Apr/22

$$Differentiatefromfirstprinciple$$$$y={log}_{x}a$$$$$$$$Mastermind$$

Commented by bobhans last updated on 29/Apr/22

$$y=\frac{\ln a}{\ln x}\Rightarrow \frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{\frac{\ln a}{\ln (x+h)}-\frac{\ln a}{\ln x}}{h}$$$$\frac{dy}{dx}=\ln a\underset{h\to 0}{\lim }\frac{\ln x-\ln (x+h)}{h\ln x\ln (x+h)}$$$$\frac{dy}{dx}=-\frac{\ln a}{\ln {}^2\left(x\right)}.\underset{h\to 0}{\lim }\frac{\ln \left(\frac{x+h}{x}\right)}{h}$$$$=-\frac{\ln a}{\ln {}^2\left(x\right)}.\underset{h\to 0}{\lim }\frac{\ln (1+\frac{h}{x})}{h}$$$$=-\frac{\ln a}{x\ln {}^2\left(x\right)}$$

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