Differentiate log(cosh(x)) + (1/2)cosh²(x) with respect to x.
Understand the Problem
The question is asking to differentiate the expression log(cosh(x)) + (1/2)cosh²(x) with respect to x. We will apply the rules of differentiation to each term separately and then combine the results.
Answer
$$ \tanh(x) + \cosh(x) \sinh(x) $$
Answer for screen readers
The derivative of the expression $\log(\cosh(x)) + \frac{1}{2}\cosh^2(x)$ with respect to $x$ is: $$ \tanh(x) + \cosh(x) \sinh(x) $$
Steps to Solve
- Differentiate the first term: $\log(\cosh(x))$
To differentiate the first term, we will use the chain rule. The derivative of $\log(u)$ is $\frac{1}{u}$ times the derivative of $u$. Here, $u = \cosh(x)$. So, $$ \frac{d}{dx} \log(\cosh(x)) = \frac{1}{\cosh(x)} \cdot \frac{d}{dx}(\cosh(x)) $$
Knowing that the derivative of $\cosh(x)$ is $\sinh(x)$, we can now compute: $$ \frac{d}{dx} \log(\cosh(x)) = \frac{\sinh(x)}{\cosh(x)} $$
This simplifies to: $$ \tanh(x) $$
- Differentiate the second term: $\frac{1}{2}\cosh^2(x)$
We will use the chain rule again. The derivative of $\frac{1}{2}u^2$ is $u \frac{du}{dx}$, where $u = \cosh(x)$. So, $$ \frac{d}{dx} \left(\frac{1}{2} \cosh^2(x)\right) = \cosh(x) \cdot \frac{d}{dx}(\cosh(x)) $$
Substituting the derivative of $\cosh(x)$: $$ = \cosh(x) \cdot \sinh(x) $$
- Combine the derivatives
Now we combine the derivatives from both terms: $$ \frac{d}{dx} \left(\log(\cosh(x)) + \frac{1}{2} \cosh^2(x)\right) = \tanh(x) + \cosh(x) \sinh(x) $$
- Final expression
So, the final result for the derivative of the given expression is: $$ \tanh(x) + \cosh(x) \sinh(x) $$
The derivative of the expression $\log(\cosh(x)) + \frac{1}{2}\cosh^2(x)$ with respect to $x$ is: $$ \tanh(x) + \cosh(x) \sinh(x) $$
More Information
The function $\tanh(x)$ is a hyperbolic function representing the ratio of the hyperbolic sine function to the hyperbolic cosine function. The term $\cosh(x) \sinh(x)$ arises often in calculus when differentiating hyperbolic functions and represents basic hyperbolic identities.
Tips
- Forgetting to apply the chain rule when differentiating composite functions like $\log(\cosh(x))$.
- Not knowing the derivatives of hyperbolic functions, particularly $\frac{d}{dx}(\cosh(x)) = \sinh(x)$ and $\frac{d}{dx}(\sinh(x)) = \cosh(x)$.
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