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 
$$
y=\sec\left(x^4\right)\cos^5 x.
$$
Using the product rule, we have
$$
\frac{d}{dx}y = \frac{d}{dx}\left[\sec\left(x^4\right)\right]\cos^5 x + \sec\left(x^4\right)\frac{d}{dx}\left[\cos^5 x\right].
$$

1. Derivative of $$ \sec\left(x^4\right) $$:

Let $$ u=x^4 $$. Then
   $$
   \frac{d}{dx}\left[\sec(u)\right] = \sec(u)\tan(u) \cdot \frac{d}{dx}(u),
   $$
   and since
   $$
   \frac{d}{dx}\left(x^4\right) = 4x^3,
   $$
   we get
   $$
   \frac{d}{dx}\left[\sec\left(x^4\right)\right] = \sec\left(x^4\right)\tan\left(x^4\right) \cdot 4x^3.
   $$

2. Derivative of $$ \cos^5 x $$:

Using the chain rule,
   $$
   \frac{d}{dx}\left[\cos^5 x\right] = 5\cos^4 x \cdot \frac{d}{dx}\left[\cos x\right],
   $$
   and since
   $$
   \frac{d}{dx}\left[\cos x\right] = -\sin x,
   $$
   we have
   $$
   \frac{d}{dx}\left[\cos^5 x\right] = -5\sin x \cos^4 x.
   $$

Putting it all together, the derivative is
$$
\frac{d}{dx}y = \left(\sec\left(x^4\right)\tan\left(x^4\right) \cdot 4x^3\right)\cos^5 x + \sec\left(x^4\right)\left(-5\sin x \cos^4 x\right).
$$
The derivative of $$ y = \sec\left(x^4\right)\cos^5 x $$ with respect to $$ x $$ is: $$ \frac{dy}{dx} = 4x^3 \sec\left(x^4\right)\tan\left(x^4\right)\cos^5 x - 5\sec\left(x^4\right)\sin x \cos^4 x. $$

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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