
The value of \[\left( {\dfrac{d}{{dx}}} \right)\left( {\dfrac{1}{{{x^4}\sec x}}} \right)\] is equal to:
A. \[\dfrac{{[4\cos x - x\sin x]}}{{{x^5}}}\]
B. \[\dfrac{{ - {{[x\sin x + 4\cos x]}^5}}}{{{x^5}}}\]
C. \[\dfrac{{[4\cos x + x\sin x]}}{{{x^5}}}\]
D. None of these
Answer
218.4k+ views
Hint:
First we will simplify the given expression by using trigonometry ratios. Then we will apply the quotient rule of derivative to find the derivative of the given expression.
Formula Used:
\[\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{\dfrac{d}{{dx}}f(x) \cdot g(x) - f(x)\dfrac{d}{{dx}}g(x)}}{{{{(g(x))}^2}}}\]
\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\].
\[\cos {\rm{ x = }}\dfrac{1}{{\sec x}}\]
Complete step-by-step answer:
We have been given the function \[\left( {\dfrac{d}{{dx}}} \right)\left( {\dfrac{1}{{{x^4}\sec x}}} \right)\]
Let \[y = \dfrac{1}{{{x^4}\sec x}}\]
Use the property of \[\cos {\rm{ x}}\]:
\[y = \dfrac{{(\cos x)}}{{{x^4}}}\]
Use quotient rule:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^4}\dfrac{d}{{dx}}\cos x - \cos x\dfrac{{d{x^4}}}{{dx}}}}{{{{\left( {{x^4}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{x^4}( - \sin x) - \cos x 4{x^3}} \right)}}{{{x^8}}}\]
We will take \[{x^3}\] common from numerator and divide it to \[{x^8}\] in denominator,
\[ = \dfrac{{ - (x\sin x + 4\cos x)}}{{{x^5}}}\]
Hence, option B is correct.
Note:
Students often make mistakes to find the derivative of \[\cos x\]. They forgot to put a negative sign of the derivative of \[\cos x\]. The derivative of \[\cos x\] is \[ - \sin x\].
First we will simplify the given expression by using trigonometry ratios. Then we will apply the quotient rule of derivative to find the derivative of the given expression.
Formula Used:
\[\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{\dfrac{d}{{dx}}f(x) \cdot g(x) - f(x)\dfrac{d}{{dx}}g(x)}}{{{{(g(x))}^2}}}\]
\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\].
\[\cos {\rm{ x = }}\dfrac{1}{{\sec x}}\]
Complete step-by-step answer:
We have been given the function \[\left( {\dfrac{d}{{dx}}} \right)\left( {\dfrac{1}{{{x^4}\sec x}}} \right)\]
Let \[y = \dfrac{1}{{{x^4}\sec x}}\]
Use the property of \[\cos {\rm{ x}}\]:
\[y = \dfrac{{(\cos x)}}{{{x^4}}}\]
Use quotient rule:
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^4}\dfrac{d}{{dx}}\cos x - \cos x\dfrac{{d{x^4}}}{{dx}}}}{{{{\left( {{x^4}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{x^4}( - \sin x) - \cos x 4{x^3}} \right)}}{{{x^8}}}\]
We will take \[{x^3}\] common from numerator and divide it to \[{x^8}\] in denominator,
\[ = \dfrac{{ - (x\sin x + 4\cos x)}}{{{x^5}}}\]
Hence, option B is correct.
Note:
Students often make mistakes to find the derivative of \[\cos x\]. They forgot to put a negative sign of the derivative of \[\cos x\]. The derivative of \[\cos x\] is \[ - \sin x\].
Recently Updated Pages
In a game two players A and B take turns in throwing class 12 maths JEE_Main

The number of ways in which 6 men and 5 women can dine class 12 maths JEE_Main

The area of an expanding rectangle is increasing at class 12 maths JEE_Main

If y xxx cdots infty then find dfracdydx A yxy 1 B class 12 maths JEE_Main

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

