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

D i f f e r e n t i a t e f r o m f i r s t p r i n c i p l e

y = {l o g}_{x} a

M a s t e r m i n d

Commented by bobhans last updated on 29/Apr/22

y = \frac{\ln a}{\ln x} \Rightarrow \frac{d y}{d x} = \lim_{h \to 0} \frac{\frac{\ln a}{\ln (x + h)} - \frac{\ln a}{\ln x}}{h}

\frac{d y}{d x} = \ln a \lim_{h \to 0} \frac{\ln x - \ln (x + h)}{h \ln x \ln (x + h)}

\frac{d y}{d x} = - \frac{\ln a}{\ln^{2} (x)} . \lim_{h \to 0} \frac{\ln (\frac{x + h}{x})}{h}

= - \frac{\ln a}{\ln^{2} (x)} . \lim_{h \to 0} \frac{\ln (1 + \frac{h}{x})}{h}

= - \frac{\ln a}{x \ln^{2} (x)}

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