Question

Solve the given differential equation \[\dfrac{d}{{dx}}(\ln x)\]?

Answer 1

Hint: For differentiation, you should know that after differentiating the term or equation given, the power of the variable is reduced by one, and differentiation of constant is always zero because derivative means measuring the change of a variable with respect to some quantity and as constant is always fixed so no change can be seen.

Complete step by step answer:
Given question is
\[\dfrac{d}{{dx}}(\ln x)\]
Let,
 \[
  Y = \ln x \\
  x = {e^{y\,}}(we\,know\,{\ln _e} = 1) \\
 \]
Now differentiating both side by x
We get,
\[\dfrac{{d(x)}}{{dx}} = \dfrac{{d({e^y})}}{{dx}} = {e^y}\]
Again differentiating again equation with respect to \[Y\] we get,
\[\dfrac{{d(Y)}}{{dy}} = \dfrac{1}{{{e^y}}}\]
And \[Y = \ln x\] , so putting value obtained from the above two equation of
\[x\,and\,Y\]
In the main equation
\[Y = \ln x\]
We get,
,\[
  \dfrac{1}{{{e^{}}}} = \dfrac{1}{{{e^{nx}}}} \\
  or\,\dfrac{1}{x} \\
 \]
Hence, Differentiation of \[\ln x\] is \[\dfrac{1}{x}\].
Formulae Used: Differentiation formulae, \[\dfrac{{d({x^n})}}{{dx}} = n \times {x^{n - 1}}\]
Additional Information: Certain differentiation identity should be learned for better understanding and fast solution, here in this case only a single term was given so it was easy to go through it, but if the equation given is long enough or two variables are given then accordingly you have to solve by assuming one variable as constant and the other one, who is also present in the derivative term should be differentiated.

Note:
Differentiation is a very easy concept until and unless the equation is given is complicated, in some cases, you have to acknowledge the basic properties of differentiation. But in most of the questions, you can go on with the basic formulae mentioned above. It is easy to understand once you practiced the question over it. Some identities are very much specified like trigonometric identity then it that case you have only the option to learn the direct derivatives.