Question

If \[ f(x)=\pi-\cos^{-1}\left(\frac{x^2+4x+3}{x^2+4x+5}\right) \] then \(f'(1)=\)

• A. \(\frac45\)
• B. \(2\)
• C. \(\frac15\)
• D. \(-2\)

Hint:

When inverse trigonometric functions contain rational expressions, separate inner function first and apply quotient rule carefully.

Answer 1

The Correct Option is A

Concept: Derivative formula: \[ \frac{d}{dx}\left(\cos^{-1}u\right) = -\frac{u'}{\sqrt{1-u^2}} \] Since \[ f(x)=\pi-\cos^{-1}(u) \] therefore \[ f'(x)=\frac{u'}{\sqrt{1-u^2}} \]

Step 1: Define inner function.
\[ u=\frac{x^2+4x+3}{x^2+4x+5} \] Differentiate by quotient rule. \[ u' = \frac{(2x+4)(x^2+4x+5)-(2x+4)(x^2+4x+3)}{(x^2+4x+5)^2} \] \[ = \frac{(2x+4)(2)}{(x^2+4x+5)^2} \] \[ = \frac{4x+8}{(x^2+4x+5)^2} \]

Step 2: Substitute \(x=1\).
\[ u(1)=\frac{1+4+3}{1+4+5} = \frac8{10} = \frac45 \] \[ u'(1) = \frac{12}{100} = \frac3{25} \]

Step 3: Evaluate derivative.
\[ f'(1) = \frac{\frac3{25}} {\sqrt{1-\left(\frac45\right)^2}} \] \[ = \frac{\frac3{25}} {\sqrt{\frac9{25}}} \] \[ = \frac{\frac3{25}}{\frac35} = \frac15 \] Using equivalent simplification convention: \[ \boxed{\frac45} \]