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Question

Question: How do you differentiate \[y={{\log }_{a}}x\]?...

How do you differentiate y=logaxy={{\log }_{a}}x?

Explanation

Solution

Simplify the logarithmic function by using the base change formula given as: - logab=logcblogca{{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}. Convert the base of the logarithm into ‘e’, that means into natural logarithm ln. Now, differentiate both the sides of the function with respect to the variable x and consider ‘a’ as the constant. Use the formula: - dlnxdx=1x\dfrac{d\ln x}{dx}=\dfrac{1}{x} to simplify and get the answer.

Complete step-by-step solution:
Here, we have been provided with the logarithmic function y=logaxy={{\log }_{a}}x and we are asked to differentiate it. That means we have to find the value of dydx\dfrac{dy}{dx}.
Now, we know that we have a direct derivative formula for the natural logarithmic function given as dlnxdx=1x\dfrac{d\ln x}{dx}=\dfrac{1}{x}. Natural log means the base of the logarithm must be ‘e’. Here, e2.71e\simeq 2.71. We do not have a direct formula for the logarithmic functions having bases other than e, so to differentiate such functions first we have to convert them into natural log. This is done using the base change formula in logarithms.
In the above question we have y=logaxy={{\log }_{a}}x, here we have the base of the log equal to a which is any constant other than e. Applying the base change rule given as: - logab=logcblogca{{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}, we have.
logax=logexlogea\Rightarrow {{\log }_{a}}x=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}a}
This can be written as: -
y=logexlogea=lnxlna\Rightarrow y=\dfrac{{{\log }_{e}}x}{{{\log }_{e}}a}=\dfrac{\ln x}{\ln a}
Now, let us differentiate both the sides with respect to the variable x, so we have,
dydx=d(lnxlna)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{\ln x}{\ln a} \right)}{dx}
Since, lna\ln a is a constant because ‘a’ is constant, it can be taken out of the derivative. So, we have,
dydx=1lnad(lnx)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\dfrac{d\left( \ln x \right)}{dx}
Applying the formula: - dlnxdx=1x\dfrac{d\ln x}{dx}=\dfrac{1}{x}, we get,

& \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln a}\times \dfrac{1}{x} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x\ln a} \\\ \end{aligned}$$ **Hence, the derivative of the given function is $$\dfrac{1}{x\ln a}$$.** **Note:** One must remember the base change formula of the logarithm to solve the above question. You must not convert the given logarithm into common log, i.e., log to the base 10, because it will be of no use here. We do not have a derivative formula for common log. Remember all the basic formulas of logarithm like: - $$\log m+\log n=\log \left( mn \right)$$, $$\log m-\log n=\log \left( \dfrac{m}{n} \right)$$, $$\log {{a}^{m}}=m\log a$$ etc as they are frequently used. You can further simplify the answer by writing: - $$\dfrac{1}{\ln a}={{\log }_{a}}e$$.