Differentiate log(sinh x + cosh x) with respect to x.
Understand the Problem
The question is asking for the derivative of the logarithm of the sum of the hyperbolic sine and cosine functions with respect to the variable x. To solve this, we will apply the chain rule and the properties of derivatives to differentiate the function.
Answer
$$ \frac{dy}{dx} = 1 $$
Answer for screen readers
The derivative of the function is:
$$ \frac{dy}{dx} = 1 $$
Steps to Solve
- Identify the function to differentiate
We need to differentiate the function $y = \ln(\sinh(x) + \cosh(x))$ with respect to $x$.
- Apply the chain rule
Using the chain rule, we know that the derivative of $y$ with respect to $x$ is given by:
$$ \frac{dy}{dx} = \frac{1}{\sinh(x) + \cosh(x)} \cdot \frac{d}{dx}(\sinh(x) + \cosh(x)) $$
- Differentiate the inner function
Now we need to differentiate the inner function $\sinh(x) + \cosh(x)$:
$$ \frac{d}{dx}(\sinh(x)) = \cosh(x) $$ $$ \frac{d}{dx}(\cosh(x)) = \sinh(x) $$
So the derivative becomes:
$$ \frac{d}{dx}(\sinh(x) + \cosh(x)) = \cosh(x) + \sinh(x) $$
- Combine the derivatives
Now we can put everything together:
$$ \frac{dy}{dx} = \frac{1}{\sinh(x) + \cosh(x)} \cdot (\sinh(x) + \cosh(x)) $$
- Simplify the expression
Notice that the $\sinh(x) + \cosh(x)$ terms cancel out:
$$ \frac{dy}{dx} = 1 $$
The derivative of the function is:
$$ \frac{dy}{dx} = 1 $$
More Information
The derivative of the logarithm of the sum of the hyperbolic sine and cosine functions simplifies nicely to 1, indicating that the function grows at a constant rate. This is interesting because it shows that the function does not vary with changes in $x$ after taking the logarithm.
Tips
- Forgetting to apply the chain rule properly can lead to incorrect derivatives.
- Not simplifying the result after differentiating, which can cause confusion.
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