Differentiate w.r.t x, \[\log \left( {\log x} \right)\]
Answer
652.2k+ views
Hint:
Use the chain rule to differentiate the given function since the function is a composite function and chain rule tells us how to differentiate a composite function which is given as
\[f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)\]
Here we need to differentiate the given function with respect to x where differentiation of expression is the rate of change of a function with respect to independent variables.
Derivation of a function is basically a measure of sensitivity to change of function value with change in the argument where argument refers to the input whose output is to be found. Derivatives are useful in finding the slope of an equation, maxima, and minima of a function when the slope is zero and is also used to check a function, whether it is increasing or decreasing.
Complete step by step solution:
Let \[y = \log (\log x) - (i)\]
Since the given function is a composite function, so we use the chain rule to differentiate equation which is given as
\[f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)\]
Hence by using the chain rule in equation (i), we can write
\[
\dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{d\left( {\log \left( {\log x} \right)} \right)}}{{dx}} \\
\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{{d\left( {\log x} \right)}}{{dx}} \\
\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{1}{x} \\
= \dfrac{1}{{x\log x}} \\
\][Since\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]]
Hence the differentiate of \[\log \left( {\log x} \right)\]with respect to x \[ = \dfrac{1}{{x\log x}}\]
The important formula used:
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]
Note:
As a constant term does not contain any variables with them when they are differentiated, then their value is zero. Derivation of a function is represented in\[\dfrac{a}{b}\], where \[a\] is the function which is being differentiated and b its independent variable by which function is being differentiated written as \[\dfrac{{dy}}{{dx}}\] where y is the function.
Use the chain rule to differentiate the given function since the function is a composite function and chain rule tells us how to differentiate a composite function which is given as
\[f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)\]
Here we need to differentiate the given function with respect to x where differentiation of expression is the rate of change of a function with respect to independent variables.
Derivation of a function is basically a measure of sensitivity to change of function value with change in the argument where argument refers to the input whose output is to be found. Derivatives are useful in finding the slope of an equation, maxima, and minima of a function when the slope is zero and is also used to check a function, whether it is increasing or decreasing.
Complete step by step solution:
Let \[y = \log (\log x) - (i)\]
Since the given function is a composite function, so we use the chain rule to differentiate equation which is given as
\[f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)\]
Hence by using the chain rule in equation (i), we can write
\[
\dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{d\left( {\log \left( {\log x} \right)} \right)}}{{dx}} \\
\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{{d\left( {\log x} \right)}}{{dx}} \\
\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{1}{x} \\
= \dfrac{1}{{x\log x}} \\
\][Since\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]]
Hence the differentiate of \[\log \left( {\log x} \right)\]with respect to x \[ = \dfrac{1}{{x\log x}}\]
The important formula used:
\[\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}\]
Note:
As a constant term does not contain any variables with them when they are differentiated, then their value is zero. Derivation of a function is represented in\[\dfrac{a}{b}\], where \[a\] is the function which is being differentiated and b its independent variable by which function is being differentiated written as \[\dfrac{{dy}}{{dx}}\] where y is the function.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

What is the maximum height of a projectile class 11 physics CBSE

Derive an expression for maximum height and range of class 11 physics CBSE

Describe various types of epithelial tissues with the class 11 biology CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

State and derive law of conservation of momentum class 11 physics CBSE

