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How do you differentiate $\dfrac{1}{\ln x}$ ?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: Differentiation is the rate change of a function with respect to the variable on which the function depends. Here, the given question can differentiate the given function in two different ways. One by chain rule and the other way by quotient rule.

Complete step by step solution:
Let us first understand what is meant by differentiation.Differentiation in mathematics can be defined in many ways. Differentiation is the rate change of a function with respect to the variable on which the function depends. If we draw a graph of the function with respect to the independent variable, then the differentiation of the function at any point is the equal to the slope of the tangent to the curve of the function at that point. If the function is f(x) then differentiation of f(x) is written as $\dfrac{d}{dx}f(x)$.

Here, the given function is $f(x)=\dfrac{1}{\ln x}$
Therefore, differentiation of $\dfrac{1}{\ln x}$ is $\dfrac{d}{dx}\left( \dfrac{1}{\ln x} \right)=\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}$.
By using chain rule, we get that
$\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}=-1{{\left( \ln x \right)}^{-2}}.\dfrac{d}{dx}\ln x$ …… (i)
The differentiation of ln(x) is equal to $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$.
Substitute this value in equation (i).
Then we get that,
$\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}=-1{{\left( \ln x \right)}^{-2}}.\dfrac{1}{x}$
Therefore, on further simplification we get that
$\therefore\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}=\dfrac{-1}{x{{\left( \ln x \right)}^{2}}}$.

Note:For the differentiation of the given function $f(x)=\dfrac{1}{\ln x}$ we can also use the product rule or quotient rule of differentiation. The product rule for the differentiation of a function $f(x)=g(x)h(x)$ is given as,
$\dfrac{d}{dx}f(x)=\left[ \dfrac{d}{dx}g(x) \right]h(x)+g(x)\left[ \dfrac{d}{dx}h(x) \right]$
The quotient rule for the differentiation of a function $f(x)=\dfrac{g(x)}{h(x)}$ is given as
$\dfrac{d}{dx}f(x)=\dfrac{\left[ \dfrac{d}{dx}g(x) \right]h(x)-g(x)\left[ \dfrac{d}{dx}h(x) \right]}{{{\left[ h(x) \right]}^{2}}}$
Therefore, we can use the quotient rule for the given function to differentiate as:
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\ln x} \right)=\dfrac{\left[ \dfrac{d}{dx}(1) \right]\ln x-1.\left[ \dfrac{d}{dx}\ln x \right]}{{{\left( \ln x \right)}^{2}}}$
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{\ln x} \right)=\dfrac{0-\left[ \dfrac{1}{x} \right]}{{{\left( \ln x \right)}^{2}}}$
$\therefore \dfrac{d}{dx}\left( \dfrac{1}{\ln x} \right)=\dfrac{-1}{x{{\left( \ln x \right)}^{2}}}$