Question:

Step 1: Establish the relationship between derivatives of inverse functions. For an inverse function $g = f^{-1}$, the derivative of $g$ at a point $y$ is given by: $$g'(y) = \frac{1}{f'(g(y))}$$ Step 2: Find the first derivative of the inverse function at $y = 3$. Since $f(0) = 3$, we have $g(3) = 0$. Therefore: $$g'(3) = \frac{1}{f'(0)} = \frac{1}{\sin^2(\sin(0+1))} = \frac{1}{\sin^2(\sin 1)}$$ Step 3: Derive the formula for the second derivative of the inverse function. Differentiating $g'(y) = \frac{1}{f'(g(y))}$ with respect to $y$ using the chain rule: $$g''(y) = -\frac{f''(g(y)) \cdot g'(y)}{[f'(g(y))]^2}$$ Step 4: Find the second derivative of $f(x)$. Starting with $f'(x) = \sin^2(\sin(x+1))$, we differentiate using the chain rule: $$f''(x) = 2\sin(\sin(x+1)) \cdot \cos(\sin(x+1)) \cdot \cos(x+1)$$ Using the double angle identity $2\sin(u)\cos(u) = \sin(2u)$: $$f''(x) = \sin(2\sin(x+1)) \cdot \cos(x+1)$$ Step 5: Evaluate $f''(x)$ at $x = 0$. Substituting $x = 0$: $$f''(0) = \sin(2\sin 1) \cdot \cos 1 = 2\sin(\sin 1)\cos(\sin 1) \cdot \cos 1$$ Step 6: Calculate $(f^{-1})''(3) = g''(3)$. Using the formula from Step 3 with $g(3) = 0$: $$g''(3) = -\frac{f''(0) \cdot g'(3)}{[f'(0)]^2}$$ Substituting the values from Steps 2 and 5: $$g''(3) = -\frac{2\sin(\sin 1)\cos(\sin 1)\cos 1 \cdot \frac{1}{\sin^2(\sin 1)}}{\sin^4(\sin 1)}$$ Simplifying: $$g''(3) = -\frac{2\cos(\sin 1)\cos 1}{\sin(\sin 1) \cdot \sin^4(\sin 1)} = -\frac{2\cos(\sin 1)\cos 1}{\sin^5(\sin 1)}$$ Therefore, $(f^{-1})''(3) = -\dfrac{2\sin(\cos 1)\sin^2 1}{\sin^6(\cos 1)}$ The answer is **Option 3**.