1. Find the following derivatives with respect to the variable as indicated (do not need to simplify) a) $$ y=\sec \left(x^{4}\right)\left(\cos ^{5} x\right) ; \frac{d}{d x} $$

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Let 
$$
f(x) = \sec(x^4) \quad \text{and} \quad g(x) = \cos^5 x.
$$

The derivative of a product is given by
$$
\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x).
$$

1. Differentiate $$ f(x)=\sec(x^4) $$:

• Recall that $$\frac{d}{du} \sec u = \sec u \tan u$$, and here $$ u = x^4 $$ so that $$ u' = 4x^3 $$.

• Thus,
$$
f'(x) = \sec(x^4)\tan(x^4)\cdot 4x^3.
$$

2. Differentiate $$ g(x)=\cos^5 x $$:

• Write $$ g(x) = (\cos x)^5 $$. Using the chain rule:
$$
g'(x) = 5(\cos x)^4 \cdot \frac{d}{dx}(\cos x),
$$
and since $$ \frac{d}{dx}(\cos x) = -\sin x $$,
$$
g'(x) = -5 \cos^4 x \sin x.
$$

3. Apply the product rule:

$$
\frac{dy}{dx} = f'(x)g(x) + f(x)g'(x).
$$

Substitute the derived expressions:

$$
\frac{dy}{dx} = \left(4x^3 \sec(x^4)\tan(x^4)\right)\cos^5 x + \sec(x^4)\left(-5 \cos^4 x \sin x\right).
$$

Thus, the derivative is:

$$
\frac{dy}{dx} = 4x^3 \sec(x^4)\tan(x^4)\cos^5 x - 5 \sec(x^4)\cos^4 x\sin x.
$$
The derivative is $$ \frac{dy}{dx} = 4x^3 \sec(x^4)\tan(x^4)\cos^5 x - 5 \sec(x^4)\cos^4 x\sin x $$.

1. Find the following derivatives with respect to the variable as indicated (do not need to simplify) a) \( y=\sec \left(x^{4}\right)\left(\cos ^{5} x\right) ; \frac{d}{d x} \)

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