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Q.Find the derivative of log⁡ex\log_e x by first principle.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2018Subjective· 6mImportance★★★★★
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Set up the difference quotient for f(x)=log⁡exf(x)=\log_e x, simplify using log properties, and use the standard limit lim⁡t→0log⁡e(1+t)t=1\lim_{t\to0}\dfrac{\log_e(1+t)}{t}=1.

By the first-principle (definition of derivative):

f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→0log⁡e(x+h)−log⁡exhf'(x) = \lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to0} \dfrac{\log_e(x+h) - \log_e x}{h}

Using the quotient rule of logarithms, log⁡e(x+h)−log⁡ex=log⁡e(x+hx)=log⁡e(1+hx)\log_e(x+h) - \log_e x = \log_e\left(\dfrac{x+h}{x}\right) = \log_e\left(1+\dfrac{h}{x}\right):

f′(x)=lim⁡h→0log⁡e(1+hx)hf'(x) = \lim_{h\to0} \dfrac{\log_e\left(1+\dfrac{h}{x}\right)}{h}

Let t=hxt = \dfrac{h}{x}, so h=xth = xt; as h→0h \to 0, t→0t \to 0 too: …

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