Find the derivative of the function, y = arctan(sinh x)

To find the derivative of the function \( y = \arctan(\sinh x) \), we will apply the chain rule and some derivative formulas.

### Step 1: Recall the derivative of \( \arctan(u) \)
The derivative of \( \arctan(u) \) is:
\[
\frac{d}{dx} \arctan(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}
\]
where \( u = \sinh(x) \) in this case.

### Step 2: Compute the derivative of \( \sinh(x) \)
We also need the derivative of \( \sinh(x) \), which is:
\[
\frac{d}{dx} \sinh(x) = \cosh(x)
\]

### Step 3: Apply the chain rule
Now apply the chain rule by substituting \( u = \sinh(x) \) into the formula for the derivative of \( \arctan(u) \):
\[
\frac{dy}{dx} = \frac{1}{1 + \sinh^2(x)} \cdot \cosh(x)
\]

### Step 4: Simplify the expression
Using the identity \( 1 + \sinh^2(x) = \cosh^2(x) \), we can simplify the expression:
\[
\frac{dy}{dx} = \frac{\cosh(x)}{\cosh^2(x)} = \frac{1}{\cosh(x)}
\]

Thus, the derivative of \( y = \arctan(\sinh(x)) \) is:
\[
\frac{dy}{dx} = \text{sech}(x)
\]
where \( \text{sech}(x) = \frac{1}{\cosh(x)} \).

### Final Answer:
\[
\frac{dy}{dx} = \text{sech}(x)
\]

Would you like more details on any of the steps or explanations?

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Here are five related questions to expand on this:

1. What is the derivative of \( y = \arctan(\cosh x) \)?
2. How do you differentiate \( y = \arcsinh(x) \)?
3. Can you compute the second derivative of \( y = \arctan(\sinh x) \)?
4. What is the Taylor series expansion of \( \arctan(\sinh x) \) around \( x = 0 \)?
5. How does the hyperbolic function derivative differ from the circular trigonometric functions?

### Tip:
Remember, hyperbolic identities like \( \cosh^2(x) - \sinh^2(x) = 1 \) can often simplify complex derivatives!

Learn how to find the derivative of y = arctan(sinh x) using the chain rule and hyperbolic identities. This step-by-step solution covers key calculus concepts, including inverse trigonometric and hyperbolic functions. Suitable for university-level calculus students.

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