Get Answers to all your Questions

header-bg qa

Find the derivative of  \tan ^{-1}\left(1+x^3\right) with respect to\left(3 x^2+2 x+1\right)

Option: 1

\frac{3 x^2}{(5 x+2)\left(x^6+2 x^3+2\right)}


Option: 2

\frac{3 x^2}{(6 x+2)\left(x^5+2 x^3+2\right)}


Option: 3

\frac{3 x^2}{(6 x+2)\left(x^6+2 x^3+2\right)}


Option: 4

\frac{3 x^2}{(6 x+2)\left(x^7+2 x^4+3\right)}


Answers (1)

best_answer

Let \hspace{0.2cm}f(x)=\tan ^{-1}\left(1+x^3\right), g(x)=3 x^2+2 x+1\\ Using \hspace{0.2cm}chain\hspace{0.2cm} and\hspace{0.2cm} power\hspace{0.2cm} rule\\, \begin{aligned} f^{\prime}(x) & =\frac{3 x^2}{x^6+2 x^3+2} \\ g^{\prime}(x) & =6 x+2 \\ \frac{f^{\prime}(x)}{g^{\prime}(x)} & =\frac{3 x^2}{(6 x+2)\left(x^6+2 x^3+2\right)} \end{aligned}

Posted by

Irshad Anwar

View full answer

JEE Main high-scoring chapters and topics

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