Question

If \[ y= \tan^{-1}\!\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\!\left(\frac{1}{x^2+3x+3}\right) + \tan^{-1}\!\left(\frac{1}{x^2+5x+7}\right), \] then \[ y'(0)= \]

• A. \(\dfrac{3}{10}\)
• B. \(-\dfrac12\)
• C. \(-\dfrac{7}{10}\)
• D. \(-\dfrac{9}{10}\) \bigskip

Hint:

For questions asking only \(y'(0)\), differentiate first and substitute \(x=0\) immediately in each term. This avoids lengthy algebraic simplification.

Answer 1

The Correct Option is D

Concept: For \(y=\tan^{-1}u\), \(y' = u'/(1+u^2)\). Differentiate each term separately, evaluate at \(x=0\), and sum.

Step 1: First term: \(u_1 = 1/(1+x+x^2)\). \(u_1' = -(1+2x)/(1+x+x^2)^2\). At \(x=0\): \(u_1=1, u_1'=-1\). \(T_1' = -1/(1+1) = -1/2\).

Step 2: Second term: \(u_2 = 1/(x^2+3x+3)\). \(u_2' = -(2x+3)/(...)^2\). At \(x=0\): \(u_2=1/3, u_2'=-1/3\). \(T_2' = (-1/3)/(1+1/9) = (-1/3)/(10/9) = -3/10\).

Step 3: Third term: \(u_3 = 1/(x^2+5x+7)\). \(u_3' = -(2x+5)/(...)^2\). At \(x=0\): \(u_3=1/7, u_3'=-5/49\). \(T_3' = (-5/49)/(1+1/49) = (-5/49)/(50/49) = -1/10\).

Step 4: Sum = \(-1/2 - 3/10 - 1/10 = -5/10 - 3/10 - 1/10 = -9/10\).

Step 5: Write the final answer. \(\boxed{-\frac{9}{10}}\)