4. $$ \left( an \left(x^{2} ight) ight)^{\prime \prime \prime} $$

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4. $$ \left(	an \left(x^{2}ight)ight)^{\prime \prime \prime} $$

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Let $$f(x)=\tan(x^2)$$. Set $$u = x^2$$ so that $$f(x)=\tan(u)$$. Then by the chain rule,
$$
f'(x)=\frac{d}{dx}\tan(u)=\sec^2(u)\,\frac{du}{dx}=\sec^2(x^2)\cdot (2x).
$$
Thus, $$f'(x)=2x\sec^2(x^2)$$.
Differentiate $$f'(x)=2x\sec^2(x^2)$$ using the product rule. Write
$$
f''(x)=\frac{d}{dx}[2x]\cdot\sec^2(x^2)+2x\cdot\frac{d}{dx}[\sec^2(x^2)].
$$
First, $$\frac{d}{dx}[2x]=2$$. For the derivative of $$\sec^2(x^2)$$, use the chain rule:
$$
\frac{d}{dx}[\sec^2(u)] = 2\sec^2(u)\tan(u)\,\frac{du}{dx}\quad \text{with}\quad u=x^2,\, \frac{du}{dx}=2x.
$$
Thus,
$$
\frac{d}{dx}[\sec^2(x^2)] = 2\sec^2(x^2)\tan(x^2)\cdot 2x = 4x\sec^2(x^2)\tan(x^2).
$$
Substitute back into the product rule:
$$
f''(x)= 2\sec^2(x^2)+ 2x\cdot [4x\sec^2(x^2)\tan(x^2)] = 2\sec^2(x^2)+8x^2\sec^2(x^2)\tan(x^2).
$$
Differentiate the second derivative
$$
f''(x)= 2\sec^2(x^2)+8x^2\sec^2(x^2)\tan(x^2).
$$
We differentiate term‐by‐term.

1. For the first term, $$2\sec^2(x^2)$$:
$$
\frac{d}{dx}[2\sec^2(x^2)] = 2\cdot \frac{d}{dx}[\sec^2(x^2)] = 2\cdot [4x\sec^2(x^2)\tan(x^2)] = 8x\sec^2(x^2)\tan(x^2).
$$

2. For the second term, $$8x^2\sec^2(x^2)\tan(x^2)$$, use the product rule. Let
$$
U(x)=8x^2 \quad \text{and} \quad V(x)=\sec^2(x^2)\tan(x^2).
$$
Then
$$
\frac{d}{dx}[U(x)V(x)] = U'(x)V(x)+U(x)V'(x).
$$
First, compute $$U'(x)=16x$$.

Now, differentiate $$V(x)=\sec^2(x^2)\tan(x^2)$$ using the product rule again. Write
$$
A(x)=\sec^2(x^2) \quad \text{and} \quad B(x)=\tan(x^2).
$$
We have already computed $$A'(x)=4x\sec^2(x^2)\tan(x^2)$$. Next, differentiate $$B(x)=\tan(x^2)$$ using the chain rule:
$$
B'(x)=\sec^2(x^2)\cdot (2x)=2x\sec^2(x^2).
$$
Thus, by the product rule for $$V(x)$$:
$$
V'(x)=A'(x)B(x)+A(x)B'(x)= [4x\sec^2(x^2)\tan(x^2)]\tan(x^2)+\sec^2(x^2)[2x\sec^2(x^2)].
$$
Simplify:
$$
V'(x)=4x\sec^2(x^2)\tan^2(x^2)+2x\sec^4(x^2).
$$
Returning to $$U(x)V(x)$$:
$$
\frac{d}{dx}[8x^2\sec^2(x^2)\tan(x^2)] = 16x\sec^2(x^2)\tan(x^2)+8x^2\Bigl[4x\sec^2(x^2)\tan^2(x^2)+2x\sec^4(x^2)\Bigr].
$$
Multiply out the second part:
$$
8x^2\cdot 4x\sec^2(x^2)\tan^2(x^2)=32x^3\sec^2(x^2)\tan^2(x^2),
$$
$$
8x^2\cdot 2x\sec^4(x^2)=16x^3\sec^4(x^2).
$$
Therefore, the derivative of the second term is
$$
16x\sec^2(x^2)\tan(x^2)+32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$

Now, combine both parts to get the third derivative:
$$
f'''(x)=8x\sec^2(x^2)\tan(x^2) + 16x\sec^2(x^2)\tan(x^2) + 32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$
Combine the like terms in the first two summands:
$$
8x\sec^2(x^2)\tan(x^2)+16x\sec^2(x^2)\tan(x^2)=24x\sec^2(x^2)\tan(x^2).
$$
Thus, the final expression for the third derivative is
$$
f'''(x)=24x\sec^2(x^2)\tan(x^2)+32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$

This can be factored slightly by extracting the common factor $$8x\sec^2(x^2)$$:
$$
f'''(x)=8x\sec^2(x^2)\Bigl[3\tan(x^2)+4x^2\tan^2(x^2)+2x^2\sec^2(x^2)\Bigr].
$$
This is a compact form of the result.
**Final Answer:** $$ f'''(x)=24x\sec^2(x^2)\tan(x^2)+32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2), $$ or equivalently, $$ f'''(x)=8x\sec^2(x^2)\Bigl[3\tan(x^2)+4x^2\tan^2(x^2)+2x^2\sec^2(x^2)\Bigr]. $$

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