Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let g(x) be the inverse of the function f(x) and <img alt="\mathrm{f^{\prime}(x)=\frac{1}{1+x^3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%5E%7B%5Cprime%7D%28x%

Let g(x) be the inverse of the function f(x) and \mathrm{f^{\prime}(x)=\frac{1}{1+x^3}}. Then \mathrm{g^{\prime}(x)} is equal to.

Option: 1

\mathrm{\frac{1}{1+(g(x))^3} }


Option: 2

\mathrm{\frac{1}{1+(f(x))^3}}


Option: 3

\mathrm{1+(g(x))^3}


Option: 4

\mathrm{1+(f(x))^3}


Answers (1)

Since g(x) is the inverse of f(x), therefore
\mathrm{ f(x)=y \Leftrightarrow g(y)=x . }
\mathrm{ g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)}, \forall x }

\mathrm{ \Rightarrow \quad g^{\prime}(f(x))=1+x^3, \forall x \\ }
\mathrm{ \Rightarrow \quad g^{\prime}(y)=1+|g(y)|^3 }             [Using f(x)=y \mathrm{\Leftrightarrow} x=g(y)]
\mathrm{ \Rightarrow \quad g^{\prime}(x)=1+|g(x)|^3 }             [Replacing y by x]

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question