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Question

Question: Find the derivative of \[{\log _{10}}x\] with respect to \[x\]....

Find the derivative of log10x{\log _{10}}x with respect to xx.

Explanation

Solution

We need to find the derivative of log10x{\log _{10}}x with respect to xx. First of all, we will simplify the given term using properties of logarithm. Using the property logab=logbloga{\log _a}b = \dfrac{{\log b}}{{\log a}}, we will first simplify the given expression i.e. log10x{\log _{10}}x. After that, we will differentiate it with respect to xx using the derivative formulas. We know, ddx(logx)=1x\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x} and ddx(c×f(x))=c(ddx(f(x)))\dfrac{d}{{dx}}\left( {c \times f(x)} \right) = c\left( {\dfrac{d}{{dx}}\left( {f(x)} \right)} \right) and so using these formulas, we will solve our problem.

Complete answer: We need to differentiate log10x{\log _{10}}x with respect to xx i.e. we need to find ddx(log10x)(1)\dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right) - - - - - (1).
For that, we will first simplify log10x{\log _{10}}x.
Using the property logab=logbloga{\log _a}b = \dfrac{{\log b}}{{\log a}}, we can write
log10x=logxlog10{\log _{10}}x = \dfrac{{\log x}}{{\log 10}}, where log10\log 10 is a constant term.
log10x=1log10×logx(2){\log _{10}}x = \dfrac{1}{{\log 10}} \times \log x - - - - - (2)
Hence, using (1) and (2), we have
ddx(log10x)=ddx(1log10×logx)\dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\log 10}} \times \log x} \right)
So, log10x=1log10×logx{\log _{10}}x = \dfrac{1}{{\log 10}} \times \log x is of the form c×f(x)c \times f(x).
Using the formula ddx(c×f(x))=c×(ddx(f(x)))\dfrac{d}{{dx}}\left( {c \times f(x)} \right) = c \times \left( {\dfrac{d}{{dx}}\left( {f(x)} \right)} \right), we have
ddx(log10x)=ddx(1log10×logx)\dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\log 10}} \times \log x} \right)
=1log10×(ddx(logx))= \dfrac{1}{{\log 10}} \times \left( {\dfrac{d}{{dx}}\left( {\log x} \right)} \right)
Now, using the formula ddx(logx)=1x\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}, we have
ddx(log10x)=ddx(1log10×logx)\dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\log 10}} \times \log x} \right)
=1log10×(ddx(logx))= \dfrac{1}{{\log 10}} \times \left( {\dfrac{d}{{dx}}\left( {\log x} \right)} \right)
=1log10×1x= \dfrac{1}{{\log 10}} \times \dfrac{1}{x}
=1xlog10= \dfrac{1}{{x\log 10}}
Therefore, we have ddx(log10x)=1xlog10\dfrac{d}{{dx}}\left( {{{\log }_{10}}x} \right) = \dfrac{1}{{x\log 10}}.
Hence, the derivative of log10x{\log _{10}}x with respect to xx is 1xlog10\dfrac{1}{{x\log 10}}.

Note:
We cannot directly just solve these types of questions. Firstly, we need to be very thorough with the logarithm properties. Also, while applying the properties, we need to make sure that we are applying the right property according to the given conditions. While differentiating, we should be careful as we usually get confused between integration and differentiation formulas. While we use the property logab=logbloga{\log _a}b = \dfrac{{\log b}}{{\log a}}, we consider the logarithmic base to be ee if anything is not given.