If y = cosh(log x) + sinh(log x), prove that y_n = 0 for n > 1.
Understand the Problem
The question requires proving that the nth derivative of the function y = cosh(log x) + sinh(log x) is zero for n > 1.Essentially, we need to demonstrate that the second derivative (and all higher-order derivatives) of the given function equals zero.
Answer
0
Answer for screen readers
The $n$th derivative of $y = cosh(log(x)) + sinh(log(x))$ for $n > 1$ is 0.
Steps to Solve
- Simplify the function
Recall the definitions of $cosh(x)$ and $sinh(x)$: $cosh(x) = \frac{e^x + e^{-x}}{2}$ and $sinh(x) = \frac{e^x - e^{-x}}{2}$.
Substitute $log(x)$ for $x$ in these definitions: $cosh(log(x)) = \frac{e^{log(x)} + e^{-log(x)}}{2}$ and $sinh(log(x)) = \frac{e^{log(x)} - e^{-log(x)}}{2}$.
Therefore, $y = cosh(log(x)) + sinh(log(x)) = \frac{e^{log(x)} + e^{-log(x)}}{2} + \frac{e^{log(x)} - e^{-log(x)}}{2} = e^{log(x)} = x$.
- Find the first derivative
$y = x$, so $\frac{dy}{dx} = 1$.
- Find the second derivative
$\frac{d^2y}{dx^2} = \frac{d}{dx}(1) = 0$.
- Find the nth derivative
Since the second derivative is 0, all higher-order derivatives (i.e., the $n$th derivative for $n > 1$) will also be 0. This is because differentiating 0 with respect to $x$ always yields 0.
The $n$th derivative of $y = cosh(log(x)) + sinh(log(x))$ for $n > 1$ is 0.
More Information
The key to this problem is recognizing and applying the definitions of $cosh(x)$ and $sinh(x)$, and simplifying the expression before differentiating. This significantly reduces the complexity of the derivatives.
Tips
A common mistake is to directly differentiate $cosh(log(x))$ and $sinh(log(x))$ without simplifying. This leads to more complex calculations and increases possibilities of making errors in differentiation. Failing to correctly apply the chain rule is another common error.
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