Question

If \(y = \sin x + e^x\), then \(\frac{d^2 x}{dy^2}\) is equal to

• A. $\frac{e^x - \sin x}{(\cos x + e^x)^2}$
• B. $\frac{e^x + \sin x}{(\cos x + e^x)^2}$
• C. $\frac{e^x - \sin x}{(\cos x + e^x)^3}$
• D. $\frac{\sin x - e^x}{(\cos x + e^x)^2}$
• E. $\frac{\sin x - e^x}{(\cos x + e^x)^3}$

Hint:

Remember: $\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$ for inverse differentiation.

Answer 1

Answer: (E)

Concept: If $y=f(x)$, then: \[ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}, \quad \frac{d^2 x}{dy^2} = -\frac{d^2 y/dx^2}{\left(\frac{dy}{dx}\right)^3} \]

Step 1: Find first derivative
\[ \frac{dy}{dx} = \cos x + e^x \]

Step 2: Find second derivative
\[ \frac{d^2 y}{dx^2} = -\sin x + e^x \]

Step 3: Apply formula
\[ \frac{d^2 x}{dy^2} = -\frac{-\sin x + e^x}{(\cos x + e^x)^3} \] \[ = \frac{\sin x - e^x}{(\cos x + e^x)^3} \] Final Conclusion:
Option (E)