Question

How do you find the first and second derivative of \[\dfrac{1}{\ln x}\]?

Answer 1

Hint: Assume the given function as \[y=\dfrac{1}{\ln x}\]. Now, differentiate both the sides with respect to x to find the first derivative, \[\dfrac{dy}{dx}\]. Apply the formula: - \[\dfrac{d\left( \dfrac{1}{v} \right)}{dx}=\left( \dfrac{-1}{{{v}^{2}}} \right)\dfrac{dv}{dx}\] to find the expression for \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]R.H.S. Here, ‘v’ is the given function of x. Use the identity: - \[\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}\] to simplify. Now, again differentiate both sides of the expression to find the second derivative, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]. Apply the formula\[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], i.e., the product rule to simplify the expression and get the answer.

Complete step by step answer:
Here, we have been provided with the function \[\dfrac{1}{\ln x}\] and we are asked to find its first and second derivative.
Now, let us assume the given function as y. So, we have,
\[\Rightarrow y=\dfrac{1}{\ln x}\]
The above function is of the form, \[y=\dfrac{1}{v}\], so differentiating both sides with respect to x, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{\ln x} \right)}{dx}\]
Applying the formula: - \[\dfrac{d\left( \dfrac{1}{v} \right)}{dx}=\left( \dfrac{-1}{{{v}^{2}}} \right)\dfrac{dv}{dx}\], we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{{{\left( \ln x \right)}^{2}}}\times \dfrac{d\left( \ln x \right)}{dx}\]
Using the identity: - \[\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}\], we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{{{\left( \ln x \right)}^{2}}}\times \dfrac{1}{x}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{x{{\left( \ln x \right)}^{2}}}\] - (1)
Therefore, we have obtained the expression for the first derivative of the function y. Now, let us find the second derivative of y. So, we have,
Differentiating the expression obtained in equation (1) with respect to x both the sides, we get,
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d\left[ \dfrac{\left( -1 \right)}{x{{\left( \ln x \right)}^{2}}} \right]}{dx}\]
Taking the constant (-1) out of the derivative in the R.H.S, we get,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( -1 \right)\times \dfrac{d}{dx}\left[ \dfrac{1}{x{{\left( \ln x \right)}^{2}}} \right]\]
Again, we can see that the function, \[\left[ \dfrac{1}{x{{\left( \ln x \right)}^{2}}} \right]\] is of the form \[y=\dfrac{1}{v}\], so we have, using the formula: - \[d\left( \dfrac{1}{v} \right)=\dfrac{-1}{{{v}^{2}}}\left( \dfrac{dv}{dx} \right)\],
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-1\times \dfrac{\left( -1 \right)}{{{\left[ x{{\left( \ln x \right)}^{2}} \right]}^{2}}}\times \dfrac{d\left[ x{{\left( \ln x \right)}^{2}} \right]}{dx} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \dfrac{d\left[ x{{\left( \ln x \right)}^{2}} \right]}{dx} \\
\end{align}\]
Here, we need to differentiate \[\left[ x{{\left( \ln x \right)}^{2}} \right]\] to get our answer. So, considering this expression as a product of two function, that is x and \[{{\left( \ln x \right)}^{2}}\] and apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], where u and v are the assumed functions of x, we get,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \left[ x\dfrac{d{{\left( \ln x \right)}^{2}}}{dx}+{{\left( \ln x \right)}^{2}}\dfrac{dx}{dx} \right]\]
Simplifying the above expression, we get,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \left[ x\times 2{{\left( \ln x \right)}^{2-1}}\dfrac{d\left( \ln x \right)}{dx}+{{\left( \ln x \right)}^{2}}\times 1 \right]\]
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \left[ \dfrac{2x\ln x}{x}+{{\left( \ln x \right)}^{2}} \right] \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \left[ 2\ln x+{{\left( \ln x \right)}^{2}} \right] \\
\end{align}\]
Taking \[\ln x\] common can cancelling the common factor, we get,
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{4}}}\times \ln x\left[ 2+\ln x \right] \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{3}}}\times \left( 2+\ln x \right) \\
\end{align}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left[ \dfrac{2+\ln x}{{{x}^{2}}{{\left( \ln x \right)}^{3}}} \right]\] - (2)
Therefore, the above expression represents the second derivative of the given function.
Hence, the expression obtained in (1) and (2) is our answers.

Note:
One may note that there are many ways to solve the question. We can convert the given logarithmic form into exponential form as \[x={{e}^{\dfrac{1}{y}}}\] and then differentiate both the sides. Once the expressions for \[\dfrac{dy}{dx}\] and \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] are obtained, we will just substitute the original value of y to get the answer. You must remember all the basic formulas of differentiation like - the chain rule, product rule, \[\dfrac{u}{v}\] rule, etc. Otherwise, it will be difficult to solve the question.