
How can one find the derivative of this function: $ {x^{\dfrac{1}{4}}} $
Answer
531.6k+ views
Hint: Always remember to assign the given function a variable, say $ y $ . When we are asked to find the derivative of a function, it means that we need to differentiate it. There are many established formulas that we can use to differentiate any given function. Always when we are asked to differentiate with respect to a variable, all the terms in the function that do not associate with that variable are to be treated as constants. Or in higher level differentiation, there are different ways of going about functions with multiple variables.
Complete step-by-step answer:
First check what the question says;
A function: $ {x^{\dfrac{1}{4}}} $ is given. And we are asked to find its derivative.
Let us assign the given function to a variable, so that we can find the derivative of that variable with respect to $ x $ . Consider the variable $ y $ as the function, it can be written as;
$ \Rightarrow y = {x^{\dfrac{1}{4}}} $
Now let us recall the general formula used to differentiate any variable, so if we are differentiating a term $ {x^n} $ then its derivative with respect to $ x $ can be written as;
$ \dfrac{d}{{dx}}({x^n}) = n \times {x^{n - 1}} $
Since in the question given to us, we have only a single term and that term is of the form $ {x^n} $ , where $ n = \dfrac{1}{4} $ , we can easily apply the formula as it is;
$ \Rightarrow \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4} \times {x^{\dfrac{1}{4} - 1}} $
Simplifying, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4} \times {x^{ - \dfrac{3}{4}}} $
$
\\
\therefore \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4}{x^{ - \dfrac{3}{4}}} \;
$
So, the correct answer is “$\dfrac{1}{4}{x^{ - \dfrac{3}{4}}}$”.
Note: Derivatives can be denoted in two ways.
Derivative of $ y $ with respect to $ x $ can be shown as: $ \dfrac{{dy}}{{dx}} $ or $ y' $ .
The various properties of derivatives are as given below:
a) The sum or difference rule: $ (m \pm n)' = m' \pm n' $
b) The product rule: $ (mn)' = m'n + n'm $
c) The quotient rule: $ (\dfrac{m}{n})' = \dfrac{{m'n - mn'}}{{{n^2}}};\;n \ne 0 $
d) Composite function (chain rule): If $ y = f(t),\;t = g(x) $ then, $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} $
The other properties describe how derivatives are treated with logarithmic, parametric, implicit functions and there are also second order derivatives.
Complete step-by-step answer:
First check what the question says;
A function: $ {x^{\dfrac{1}{4}}} $ is given. And we are asked to find its derivative.
Let us assign the given function to a variable, so that we can find the derivative of that variable with respect to $ x $ . Consider the variable $ y $ as the function, it can be written as;
$ \Rightarrow y = {x^{\dfrac{1}{4}}} $
Now let us recall the general formula used to differentiate any variable, so if we are differentiating a term $ {x^n} $ then its derivative with respect to $ x $ can be written as;
$ \dfrac{d}{{dx}}({x^n}) = n \times {x^{n - 1}} $
Since in the question given to us, we have only a single term and that term is of the form $ {x^n} $ , where $ n = \dfrac{1}{4} $ , we can easily apply the formula as it is;
$ \Rightarrow \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4} \times {x^{\dfrac{1}{4} - 1}} $
Simplifying, we get:
$ \Rightarrow \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4} \times {x^{ - \dfrac{3}{4}}} $
$
\\
\therefore \dfrac{d}{{dx}}({x^{\dfrac{1}{4}}}) = \dfrac{1}{4}{x^{ - \dfrac{3}{4}}} \;
$
So, the correct answer is “$\dfrac{1}{4}{x^{ - \dfrac{3}{4}}}$”.
Note: Derivatives can be denoted in two ways.
Derivative of $ y $ with respect to $ x $ can be shown as: $ \dfrac{{dy}}{{dx}} $ or $ y' $ .
The various properties of derivatives are as given below:
a) The sum or difference rule: $ (m \pm n)' = m' \pm n' $
b) The product rule: $ (mn)' = m'n + n'm $
c) The quotient rule: $ (\dfrac{m}{n})' = \dfrac{{m'n - mn'}}{{{n^2}}};\;n \ne 0 $
d) Composite function (chain rule): If $ y = f(t),\;t = g(x) $ then, $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} $
The other properties describe how derivatives are treated with logarithmic, parametric, implicit functions and there are also second order derivatives.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

