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
\[
y'(0)=
\]
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
\[
y'(0)=
\]
Show Hint
For questions asking only \(y'(0)\), differentiate first and substitute \(x=0\) immediately in each term. This avoids lengthy algebraic simplification.
Updated On: Jul 9, 2026
- \(\dfrac{3}{10}\)
- \(-\dfrac12\)
- \(-\dfrac{7}{10}\)
- \(-\dfrac{9}{10}\) \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For
\[
y=\tan^{-1}(u),
\]
\[
\frac{dy}{dx}
=
\frac{u'}{1+u^2}.
\]
We differentiate each term separately and then substitute \(x=0\).
Step 1: Differentiate the first term. Let \[ u_1=\frac1{1+x+x^2}. \] Then \[ u_1' = -\frac{1+2x}{(1+x+x^2)^2}. \] Hence \[ \frac{d}{dx}\tan^{-1}(u_1) = \frac{u_1'}{1+u_1^2}. \] At \(x=0\), \[ u_1=1, \qquad u_1'=-1. \] Therefore \[ T_1' = \frac{-1}{1+1} = -\frac12. \]
Step 2: Differentiate the second term. Let \[ u_2=\frac1{x^2+3x+3}. \] Then \[ u_2' = -\frac{2x+3}{(x^2+3x+3)^2}. \] At \(x=0\), \[ u_2=\frac13, \qquad u_2'=-\frac13. \] Thus \[ T_2' = \frac{-\frac13} {1+\frac19} = -\frac13\cdot\frac9{10} = -\frac3{10}. \]
Step 3: Differentiate the third term. Let \[ u_3=\frac1{x^2+5x+7}. \] Then \[ u_3' = -\frac{2x+5}{(x^2+5x+7)^2}. \] At \(x=0\), \[ u_3=\frac17, \qquad u_3'=-\frac5{49}. \] Hence \[ T_3' = \frac{-\frac5{49}} {1+\frac1{49}} = -\frac5{50} = -\frac1{10}. \]
Step 4: Add the derivatives. \[ y'(0) = T_1'+T_2'+T_3'. \] \[ = -\frac12-\frac3{10}-\frac1{10}. \] \[ = -\frac5{10}-\frac3{10}-\frac1{10}. \] \[ = -\frac9{10}. \]
Step 5: Write the final answer. \[ \boxed{-\frac9{10}} \]
Step 1: Differentiate the first term. Let \[ u_1=\frac1{1+x+x^2}. \] Then \[ u_1' = -\frac{1+2x}{(1+x+x^2)^2}. \] Hence \[ \frac{d}{dx}\tan^{-1}(u_1) = \frac{u_1'}{1+u_1^2}. \] At \(x=0\), \[ u_1=1, \qquad u_1'=-1. \] Therefore \[ T_1' = \frac{-1}{1+1} = -\frac12. \]
Step 2: Differentiate the second term. Let \[ u_2=\frac1{x^2+3x+3}. \] Then \[ u_2' = -\frac{2x+3}{(x^2+3x+3)^2}. \] At \(x=0\), \[ u_2=\frac13, \qquad u_2'=-\frac13. \] Thus \[ T_2' = \frac{-\frac13} {1+\frac19} = -\frac13\cdot\frac9{10} = -\frac3{10}. \]
Step 3: Differentiate the third term. Let \[ u_3=\frac1{x^2+5x+7}. \] Then \[ u_3' = -\frac{2x+5}{(x^2+5x+7)^2}. \] At \(x=0\), \[ u_3=\frac17, \qquad u_3'=-\frac5{49}. \] Hence \[ T_3' = \frac{-\frac5{49}} {1+\frac1{49}} = -\frac5{50} = -\frac1{10}. \]
Step 4: Add the derivatives. \[ y'(0) = T_1'+T_2'+T_3'. \] \[ = -\frac12-\frac3{10}-\frac1{10}. \] \[ = -\frac5{10}-\frac3{10}-\frac1{10}. \] \[ = -\frac9{10}. \]
Step 5: Write the final answer. \[ \boxed{-\frac9{10}} \]
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