Question

How do you find the derivative of $\dfrac{1}{{\log x}}$?

Answer 1

Here we will rearrange the given expression and use the chain rule to find the derivative of the equation. On doing some simplification we get the required answer.

Formula used: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step-by-step solution:
We have the given equation as:
$\Rightarrow \dfrac{d}{{dx}}\dfrac{1}{{\log x}}$
Now since the term $\log x$ is in the denominator of the equation, we can write the term in the numerator in the inverse format as:
$\Rightarrow \dfrac{d}{{dx}}{(\log x)^{ - 1}}$
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule by writing the term as:
$\Rightarrow \dfrac{d}{{dx}}{(\log x)^{ - 1}}$
Now we know that the formula for the chain rule is: $F'(x) = f'(g(x))g'(x)$
Since we have to use the chain rule on the given expression, we will write it as:
$\Rightarrow \dfrac{d}{{dx}}\left( {{{(\log x)}^{ - 1}}\dfrac{d}{{dx}}(\log x)} \right)$
On derivative we get,
$\Rightarrow - 1 \times {(\log x)^{ - 2}} \times \dfrac{d}{{dx}}(\log x)$
Since the term in the above equation is in the form of negative power, we can invert and write it as:
$\Rightarrow - 1 \times \dfrac{1}{{{{(\log x)}^2}}} \times \dfrac{d}{{dx}}(\log x)$
Now we know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
We can write the above expression as:
$\Rightarrow - 1 \times \dfrac{1}{{{{(\log x)}^2}}} \times \dfrac{1}{x}$
On simplifying and writing the equation we get:
$\Rightarrow \dfrac{{ - 1}}{{x{{(\log x)}^2}}}$

Therefore$\Rightarrow \dfrac{d}{{dx}}\dfrac{1}{{\log x}} = \dfrac{{ - 1}}{{x{{(\log x)}^2}}}$.

Note: For these types of questions the formulas for the derivatives of the terms should be remembered.
It is to be remembered that the inverse of the number, also called the reciprocal of the number is the number diving $1$, for example the reciprocal of $a$ is $\dfrac{1}{a}$, and it can also be expressed in terms of power as ${a^{ - 1}}$.
The inverse of the derivative is the integration and vice versa. If the derivative of a term $a$ is $b$, then the integration of the term $b$ will be $a$.