Differentiate tan^{-1}(sinh(x²)) with respect to x².

Steps to Solve

1. Identify the function and the need for the chain rule

We want to find the derivative of the function $y = \tan^{-1}(\sinh(x^2))$ with respect to $x^2$. This means we will apply the chain rule.

2. Write the derivative of the outer function

The derivative of the inverse tangent function $\tan^{-1}(u)$, where $u = \sinh(x^2)$, is given by:

$$ \frac{dy}{du} = \frac{1}{1 + u^2} $$

So we have:

$$ \frac{dy}{du} = \frac{1}{1 + \sinh^2(x^2)} $$

3. Write the derivative of the inner function

Next, we need to find the derivative of $u = \sinh(x^2)$ with respect to $x^2$. The derivative of $\sinh(v)$ with respect to $v$ is:

$$ \frac{du}{dv} = \cosh(v) $$

Where $v = x^2$, we have:

$$ \frac{du}{dx^2} = \cosh(x^2) $$

4. Apply the chain rule

According to the chain rule:

$$ \frac{dy}{dx^2} = \frac{dy}{du} \cdot \frac{du}{dx^2} $$

Substituting the values we found:

$$ \frac{dy}{dx^2} = \frac{1}{1 + \sinh^2(x^2)} \cdot \cosh(x^2) $$

5. Simplify the expression

To finalize, we simplify the expression:

$$ \frac{dy}{dx^2} = \frac{\cosh(x^2)}{1 + \sinh^2(x^2)} $$

Using the identity $1 + \sinh^2(v) = \cosh^2(v)$, we can rewrite:

$$ \frac{dy}{dx^2} = \frac{\cosh(x^2)}{\cosh^2(x^2)} = \frac{1}{\cosh(x^2)} $$