Question Statement

Find $\frac{dy}{dx}$ given:

$y=\frac{1}{{\tan }^{-1}(x^2)}$

---

Background and Explanation

This problem requires differentiating a composite function involving the inverse tangent function. You will need to apply the chain rule multiple times: once for the power rule (since the function is raised to the power $-1$) and again for the inverse tangent function itself.

---

Solution

First, rewrite the function using a negative exponent to make the differentiation clearer:

$y={\left[{\tan }^{-1}(x^2)\right]}^{-1}$

Now differentiate with respect to $x$ using the chain rule and power rule:

Step 1: Apply the power rule to the outer function $[\cdot {]}^{-1}$

$\frac{dy}{dx}=(-1){\left[{\tan }^{-1}(x^2)\right]}^{-1-1}\cdot \frac{d}{dx}\left({\tan }^{-1}(x^2)\right)$

$\frac{dy}{dx}=(-1){\left({\tan }^{-1}(x^2)\right)}^{-2}\cdot \frac{d}{dx}\left({\tan }^{-1}(x^2)\right)$

Step 2: Differentiate the inner function ${\tan }^{-1}(x^2)$ using the chain rule

Recall that $\frac{d}{du}({\tan }^{-1}u)=\frac{1}{1+u^2}$. Here $u=x^2$, so:

$\frac{d}{dx}\left({\tan }^{-1}(x^2)\right)=\frac{1}{1+(x^2{)}^2}\cdot \frac{d}{dx}(x^2)$

$=\frac{1}{1+x^4}\cdot 2x$

Step 3: Combine the results

Substituting back into our expression from Step 1:

$\frac{dy}{dx}=\frac{-1}{{\left({\tan }^{-1}(x^2)\right)}^2}\cdot \frac{1}{1+x^4}\cdot 2x$

Step 4: Simplify the final expression

$\frac{dy}{dx}=\frac{-2x}{{\left({\tan }^{-1}(x^2)\right)}^2(1+x^4)}$

Or equivalently:

$\boxed{\frac{dy}{dx}=-\frac{2x}{({\tan }^{-1}(x^2){)}^2(1+x^4)}}$

---

Key Formulas or Methods Used

• Power Rule: $\frac{d}{dx}[u^n]=n\cdot u^{n-1}\cdot \frac{du}{dx}$
• Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$
• Derivative of Inverse Tangent: $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$ or generally $\frac{d}{dx}({\tan }^{-1}u)=\frac{1}{1+u^2}\cdot \frac{du}{dx}$
• Power Rule for $x^n$: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

---

Summary of Steps

1. Rewrite $y=\frac{1}{{\tan }^{-1}(x^2)}$ as $y=[{\tan }^{-1}(x^2){]}^{-1}$ to prepare for power rule differentiation
2. Apply the power rule: bring down $-1$ and reduce exponent by $1$, then multiply by derivative of inner function
3. Differentiate ${\tan }^{-1}(x^2)$ using chain rule: $\frac{1}{1+(x^2{)}^2}\cdot 2x=\frac{2x}{1+x^4}$
4. Multiply all components together and simplify signs and exponents to get final result: $\frac{-2x}{({\tan }^{-1}(x^2){)}^2(1+x^4)}$