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Question

Question: What is the domain of the derivative of \[\ln x\] ?...

What is the domain of the derivative of lnx\ln x ?

Explanation

Solution

Domain of a function is the set of all the input values upon which the function is defined. In this, we need to find the domain of the derivative of lnx\ln x. For that, we will first find the derivative of lnx\ln x. After that, we will try to find the values for which the derivative of lnx\ln x is not defined. We will then subtract the points for which the derivative is not defined from the set of real numbers to find the domain of that function.

Complete step by step answer:
We need to find the domain of the derivative of lnx\ln x.
First of all, we will find the derivative of lnx\ln x
Let y=lnxy = \ln x. So, we have to find dydx\dfrac{{dy}}{{dx}}.
Differentiating both sides of y=lnxy = \ln x, we get
dydx=ddx(lnx)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\ln x} \right)
Using ddx(lnx)=1x\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}, we get
dydx=ddx(lnx)=1x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x}
Hence, we get
dydx=1x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{x}
Hence, the derivative of lnx\ln x is 1x\dfrac{1}{x}.

Now, we need to find the domain of derivatives of lnx\ln x. i.e. we need to find the domain of 1x\dfrac{1}{x}.
Let us first find the points where 1x\dfrac{1}{x} is not defined.
We know that a fraction ab\dfrac{a}{b} is not defined on the points where b=0b = 0.
Hence, 1x\dfrac{1}{x} is not defined for x=0x = 0.
So, we see that 1x\dfrac{1}{x} is not defined only when x=0x = 0 and is defined for all the other real values of xx.
Hence, Domain of 1x\dfrac{1}{x} is \mathbb{R} - \left\\{ 0 \right\\}, where R\mathbb{R} is the set of all real numbers.

Therefore, we get, Domain of the derivative of lnx\ln x is \mathbb{R} - \left\\{ 0 \right\\}, where R\mathbb{R} is the set of all real numbers.

Note: To find the domain of a fraction f(x)g(x)\dfrac{{f\left( x \right)}}{{g\left( x \right)}}, we usually solve g(x)=0g\left( x \right) = 0 and then delete all the values of xx which satisfies g(x)=0g\left( x \right) = 0 from the set of Real Numbers. In case we forget the formula for the derivative of lnx\ln x, we can use the First Principle of Differentiation. According to the First Principle of Differentiation, f(x)=limh0f(x+h)f(x)hf'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}, where f(x)f'\left( x \right) is the derivative of the function f(x)f\left( x \right) with respect to xx.