Differentiate log(cosh(x)) + (1/2 cosh²(x)) with respect to x.
Understand the Problem
The question is asking us to differentiate the function log(cosh(x)) + (1/2 cosh²(x)) with respect to x. We will apply rules of differentiation including the chain rule and the quotient rule to find the derivative.
Answer
$$ \tanh(x) + \cosh(x) \sinh(x) $$
Answer for screen readers
The derivative of the function $\log(\cosh(x)) + \frac{1}{2} \cosh^2(x)$ with respect to $x$ is:
$$ \tanh(x) + \cosh(x) \sinh(x) $$
Steps to Solve
- Differentiate the first term
We need to differentiate the function $\log(\cosh(x))$.
The derivative of $\log(u)$, where $u = \cosh(x)$, is given by:
$$ \frac{d}{dx} \log(u) = \frac{1}{u} \cdot \frac{du}{dx} $$
where $\frac{du}{dx} = \sinh(x)$. Thus, we have:
$$ \frac{d}{dx} \log(\cosh(x)) = \frac{1}{\cosh(x)} \cdot \sinh(x) $$
- Differentiate the second term
Now we differentiate the term $\frac{1}{2} \cosh^2(x)$.
Using the chain rule, we find:
$$ \frac{d}{dx} \left(\frac{1}{2} \cosh^2(x)\right) = \frac{1}{2} \cdot 2 \cosh(x) \cdot \sinh(x) = \cosh(x) \sinh(x) $$
- Combine the derivatives
Now, we combine the derivatives from both steps:
The overall derivative is given by:
$$ \frac{d}{dx} \log(\cosh(x)) + \frac{d}{dx} \left(\frac{1}{2} \cosh^2(x)\right) $$
Substituting the results from steps 1 and 2:
$$ \frac{\sinh(x)}{\cosh(x)} + \cosh(x) \sinh(x) $$
- Simplify the expression
We can express the overall derivative in a simplified form. Notice that $\frac{\sinh(x)}{\cosh(x)}$ is equal to $\tanh(x)$. So, the final derivative becomes:
$$ \tanh(x) + \cosh(x) \sinh(x) $$
The derivative of the function $\log(\cosh(x)) + \frac{1}{2} \cosh^2(x)$ with respect to $x$ is:
$$ \tanh(x) + \cosh(x) \sinh(x) $$
More Information
The derivative of hyperbolic functions can often be simplified using trigonometric identities. Here, the use of $\tanh(x)$, which is related to both $\sinh(x)$ and $\cosh(x)$, simplifies our final answer.
Tips
- Forgetting to apply the chain rule correctly when differentiating composite functions, such as $\cosh^2(x)$.
- Confusing $\sinh(x)$ and $\cosh(x)$ derivatives; remember that $\frac{d}{dx}(\sinh(x)) = \cosh(x)$ and $\frac{d}{dx}(\cosh(x)) = \sinh(x)$.
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