Differentiate tan-1(sinh x²) with respect to x²

Steps to Solve

1. Define the function

Let $u = x^2$. Then the function becomes $\tan^{-1}(\sinh(u))$. We need to find $\frac{d}{du} \tan^{-1}(\sinh(u))$.

2. Apply the chain rule

The derivative of $\tan^{-1}(v)$ with respect to $v$ is $\frac{1}{1+v^2}$. Therefore, $\frac{d}{du} \tan^{-1}(\sinh(u)) = \frac{1}{1 + (\sinh(u))^2} \cdot \frac{d}{du} \sinh(u)$.

3. Differentiate $\sinh(u)$

The derivative of $\sinh(u)$ with respect to $u$ is $\cosh(u)$. So, $\frac{d}{du} \sinh(u) = \cosh(u)$.

4. Substitute the derivative of $\sinh(u)$

Substituting this back into the expression from step 2, we get: $\frac{1}{1 + \sinh^2(u)} \cdot \cosh(u) = \frac{\cosh(u)}{1 + \sinh^2(u)}$.

5. Simplify using the identity $\cosh^2(u) - \sinh^2(u) = 1$

We know that $\cosh^2(u) - \sinh^2(u) = 1$, which implies $1 + \sinh^2(u) = \cosh^2(u)$.

6. Final simplification

Substitute $1 + \sinh^2(u)$ with $\cosh^2(u)$ in the expression: $\frac{\cosh(u)}{\cosh^2(u)} = \frac{1}{\cosh(u)}$.