If y = cosh(log x) + sinh(log x), prove that...

Steps to Solve

1. Express y in terms of exponentials

Recall the definitions of $\cosh(u)$ and $\sinh(u)$: $\cosh(u) = \frac{e^u + e^{-u}}{2}$ and $\sinh(u) = \frac{e^u - e^{-u}}{2}$

Substitute $u = \log x$: $y = \cosh(\log x) + \sinh(\log x) = \frac{e^{\log x} + e^{-\log x}}{2} + \frac{e^{\log x} - e^{-\log x}}{2}$

2. Simplify the expression for y

Simplify the expression by combining like terms: $y = \frac{e^{\log x} + e^{-\log x} + e^{\log x} - e^{-\log x}}{2} = \frac{2e^{\log x}}{2} = e^{\log x} = x$

3. Differentiate y with respect to x (first derivative)

Differentiate $y = x$ with respect to $x$: $\frac{dy}{dx} = 1$

4. Differentiate again with respect to x (second derivative)

Differentiate $\frac{dy}{dx} = 1$ with respect to $x$: $\frac{d^2y}{dx^2} = 0$

5. Deduce the solution / Conclusion Given that $\frac{d^2y}{dx^2} = 0$, and $\frac{dy}{dx} = 1$, we can say any expression based on these values can be found. Since the question is cut off we can assume that we have to find the value for the expression: $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx}$. $x^2 (0) + x (1) = x$.