Let fx=tan^(-1)(x^3-1)/(x^2+x), then the value of 17f^(′) 2 is equal to

Correct answer: 7.4. fx=tan^(-1)(x^2-1)/x1+x^21/x=tan^(-1)x^2-tan^(-1)1/x f^(′) x =2x/(1+x^4) +1/(1+ 1 x^2) 1/(x^2) ⇒2x/(1+x^4) +1/(1+x^2) f^(')2=4/17+1/5⇒17f^(')2=4+17/5

Q28 / 46 · NTA Abhyas (JEE Main) 2020 · Mathematics · Differentiation

Let $f (x) = t a n^{- 1} (\frac{x^{3} - 1}{x^{2} + x})$ , then the value of $17 f^{'} (2)$ is equal to

Correct answer: 7.4

Solution

f (x) = t a n^{- 1} \{\frac{x^{2} - \frac{1}{x}}{1 + x^{2} \frac{1}{x}}\} = t a n^{- 1} x^{2} - t a n^{- 1} (\frac{1}{x})

f^{'} (x) = \frac{2 x}{1 + x^{4}} + \frac{1}{(1 + \frac{1}{x^{2}})} \frac{1}{x^{2}} \Rightarrow \frac{2 x}{1 + x^{4}} + \frac{1}{1 + x^{2}}

f^{'} (2) = \frac{4}{17} + \frac{1}{5} \Rightarrow 17 f^{'} (2) = 4 + \frac{17}{5}

Hence, the correct answer is 7.4.

Let fx=tan-1x3-1x2+x, then the value of 17 f ′ 2 is equal to

Solution

More Differentiation Questions

Let $f (x) = t a n^{- 1} (\frac{x^{3} - 1}{x^{2} + x})$ , then the value of $17 f^{'} (2)$ is equal to