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Chapter 6: Problem 3

Find the derivative of \(y=f(x)=\log _{a} x\), using first principle.

Short Answer

Expert verified
The derivative of \(y=f(x)=\log_{a}x\) is \(f'(x)=\frac{1}{x\ln{a}}\).

Step by step solution

01

Express the function in terms of natural logarithms

Typically, differentiation becomes simpler when the function is expressed as a natural logarithm. To achieve this, use the change of base formula, which states that \(\log_{b}a = \frac{\ln a}{\ln b}\). So, rewrite the given function \(y=f(x)=\log_{a}x\) as \(y=f(x)=\frac{\ln x}{\ln a}\).
02

Application of first principle

The first principle of differentiation states that the derivative of \(f(x)\) with respect to \(x\), denoted \(f'(x)\) or \(\frac{df}{dx}\), is given as \(\lim_{{h}\rightarrow{0}}\frac{f(x+h)-f(x)}{h}\). Plug the function from Step 1 into this principle, which results in \(\frac{df}{dx}=\lim_{h\rightarrow 0}\frac{\frac{\ln (x+h)}{\ln a}-\frac{\ln x}{\ln a}}{h}\).
03

Simplify the expression

Simplify the expression in the limit. Use properties of logarithms, specifically the subtraction rule (which states that \(\log{b} - \log{a} = \log{\frac{b}{a}}\)), to combine the logarithms in the numerator. This results in \(\frac{df}{dx}=\lim_{h\rightarrow 0}\frac{\ln{\frac{x+h}{x}}}{h \cdot \ln{a}}\).
04

Further simplify using limits

Apply the fact that \(\lim_{h\rightarrow 0}\frac{\ln (1+h)}{h}=1\). You must express \(\frac{\ln{\frac{x+h}{x}}}{h}\) as \(\ln (1 + \frac{h}{x})\). Once formatted as such, the limit can be applied. We end up with \(\frac{df}{dx}=\frac{1}{x \cdot \ln{a}}\) which is the derivative.

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Most popular questions from this chapter

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