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

Question

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

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We start with the function

$$
f(x)=\tan(x^2),
$$

and set $$ u=x^2 $$, so that $$ f(x)=\tan(u) $$.

The first derivative is obtained using the chain rule:

$$
f'(x)=\sec^2(u)\cdot u'(x) \quad \text{with} \quad u'(x)=2x.
$$

Thus,

$$
f'(x)=2x\sec^2(x^2).
$$
Differentiate $$ f'(x)=2x\sec^2(x^2) $$ with respect to $$ x $$ using the product rule:

$$
\frac{d}{dx}[2x\sec^2(x^2)] = 2\frac{d}{dx}(x\sec^2(x^2)).
$$

Let $$ g(x)=x $$ and $$ h(x)=\sec^2(x^2) $$.

First, $$ g'(x)=1 $$. Next, differentiate $$ h(x) $$ using the chain rule:

$$
h'(x)=\frac{d}{dx}\left[\sec^2(x^2)\right] = 2\sec^2(x^2)\tan(x^2)\cdot\frac{d}{dx}(x^2)=4x\sec^2(x^2)\tan(x^2).
$$

Now, applying the product rule:

$$
\frac{d}{dx}(x\sec^2(x^2)) = g'(x)h(x)+g(x)h'(x)=\sec^2(x^2)+x\cdot\left(4x\sec^2(x^2)\tan(x^2)\right).
$$

Thus,

$$
f''(x)=2\left[\sec^2(x^2)+4x^2\sec^2(x^2)\tan(x^2)\right]=2\sec^2(x^2)+8x^2\sec^2(x^2)\tan(x^2).
$$
We now differentiate $$ f''(x)=2\sec^2(x^2)+8x^2\sec^2(x^2)\tan(x^2) $$ term by term.

$$ \textbf{First Term:} $$

Differentiate $$ 2\sec^2(x^2) $$:

$$
\frac{d}{dx}\left[2\sec^2(x^2)\right] = 2\cdot\frac{d}{dx}\left[\sec^2(x^2)\right]=2\cdot\left(2\sec^2(x^2)\tan(x^2)\cdot2x\right)=8x\sec^2(x^2)\tan(x^2).
$$

$$ \textbf{Second Term:} $$

Differentiate $$ 8x^2\sec^2(x^2)\tan(x^2) $$ using the product rule. Let

$$
G(x)=8x^2 \quad \text{and} \quad H(x)=\sec^2(x^2)\tan(x^2).
$$

We have $$ G'(x)=16x $$.

Now, differentiate $$ H(x)=\sec^2(x^2)\tan(x^2) $$ using the product rule:

$$
H'(x)=\frac{d}{dx}\left[\sec^2(x^2)\right]\tan(x^2)+\sec^2(x^2)\frac{d}{dx}[\tan(x^2)].
$$

Previously, we found:

$$
\frac{d}{dx}\left[\sec^2(x^2)\right]=4x\sec^2(x^2)\tan(x^2), \quad \frac{d}{dx}[\tan(x^2)]=2x\sec^2(x^2).
$$

Thus,

$$
H'(x)=4x\sec^2(x^2)\tan(x^2)\cdot\tan(x^2) + \sec^2(x^2)\cdot2x\sec^2(x^2) 
=4x\sec^2(x^2)\tan^2(x^2)+2x\sec^4(x^2).
$$

Now, by the product rule on $$ G(x)H(x) $$:

$$
\frac{d}{dx}[G(x)H(x)] = G'(x)H(x)+G(x)H'(x).
$$

Substitute the expressions:

$$
= 16x\,\sec^2(x^2)\tan(x^2) + 8x^2\left(4x\sec^2(x^2)\tan^2(x^2)+2x\sec^4(x^2)\right).
$$

Simplify the second part:

$$
8x^2\cdot4x\sec^2(x^2)\tan^2(x^2)=32x^3\sec^2(x^2)\tan^2(x^2),
$$

$$
8x^2\cdot2x\sec^4(x^2)=16x^3\sec^4(x^2).
$$

Therefore, the derivative of the second term is:

$$
16x\sec^2(x^2)\tan(x^2) + 32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$

Adding the derivative of the first term, the third derivative is:

$$
f'''(x)=8x\sec^2(x^2)\tan(x^2) + 16x\sec^2(x^2)\tan(x^2) + 32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$

Combine like terms:

$$
8x\sec^2(x^2)\tan(x^2)+16x\sec^2(x^2)\tan(x^2)=24x\sec^2(x^2)\tan(x^2).
$$

Thus, the consolidated form is:

$$
f'''(x)=24x\sec^2(x^2)\tan(x^2)+32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2).
$$
We can also factor out common terms to present the answer more compactly. Notice that every term has a factor of $$8x\sec^2(x^2)$$:

$$
f'''(x)=8x\sec^2(x^2)\Bigl[3\tan(x^2)+4x^2\tan^2(x^2)+2x^2\sec^2(x^2)\Bigr].
$$

This factorization provides a neat final answer.
**Final Answer:** The third derivative of $$ \tan(x^2) $$ is given by $$ f'''(x)=24x\sec^2(x^2)\tan(x^2)+32x^3\sec^2(x^2)\tan^2(x^2)+16x^3\sec^4(x^2), $$ 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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