1.5 Differentiate the following: $$ 1.5 .1 y=x^{4}\left(4-x^{2}\right)^{\frac{1}{2}} $$ $$ 1.5 .2 y=\frac{\cos x}{x^{2}} $$ $$ 1.5 .3 y=\left(1-x^{3}\right) e^{3} x $$

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1.5.1  
Let  
$$
y = x^4\left(4-x^2\right)^{\frac{1}{2}}.
$$  
Define  
$$
u = x^4,\quad v = \left(4-x^2\right)^{\frac{1}{2}}.
$$  
Then  
$$
u' = 4x^3,
$$
and using the chain rule for $$ v $$:
$$
v' = \frac{1}{2}\left(4-x^2\right)^{-\frac{1}{2}} \cdot (-2x) = -\frac{x}{\left(4-x^2\right)^{\frac{1}{2}}}.
$$
By the product rule:
$$
y' = u'v + uv' = 4x^3\left(4-x^2\right)^{\frac{1}{2}} - \frac{x^5}{\left(4-x^2\right)^{\frac{1}{2}}}.
$$
Factor out $$\frac{x^3}{\left(4-x^2\right)^{\frac{1}{2}}}$$:
$$
y' = \frac{x^3}{\left(4-x^2\right)^{\frac{1}{2}}}\left[4\left(4-x^2\right)-x^2\right].
$$
Simplify the bracket:
$$
4\left(4-x^2\right)-x^2 = 16 - 4x^2 - x^2 = 16 - 5x^2.
$$
Thus, the derivative is:
$$
y' = \frac{x^3\left(16-5x^2\right)}{\left(4-x^2\right)^{\frac{1}{2}}}.
$$

1.5.2  
Given  
$$
y = \frac{\cos x}{x^2},
$$  
rewrite as  
$$
y = \cos x\cdot x^{-2}.
$$
Differentiate using the product rule, where
$$
u = \cos x,\quad v = x^{-2}.
$$
Then,
$$
u' = -\sin x,\quad v' = -2x^{-3}.
$$
Thus,
$$
y' = u'v + uv' = \left(-\sin x\right)x^{-2} + \cos x\left(-2x^{-3}\right) = -\frac{\sin x}{x^2} - \frac{2\cos x}{x^3}.
$$

1.5.3  
Assuming the function is  
$$
y = (1-x^3)e^{3x},
$$
let
$$
u = 1-x^3,\quad v = e^{3x}.
$$
Differentiate each:
$$
u' = -3x^2,
$$
and using the chain rule,
$$
v' = 3e^{3x}.
$$
By the product rule:
$$
y' = u'v + uv' = \left(-3x^2\right)e^{3x} + (1-x^3)\left(3e^{3x}\right).
$$
Factor out $$3e^{3x}$$:
$$
y' = 3e^{3x}\left[(1-x^3) - x^2\right] = 3e^{3x}\left(1-x^2-x^3\right).
$$
1.5.1 $$ y' = \frac{x^3(16 - 5x^2)}{\sqrt{4 - x^2}} $$ 1.5.2 $$ y' = -\frac{\sin x}{x^2} - \frac{2\cos x}{x^3} $$ 1.5.3 $$ y' = 3e^{3x}(1 - x^2 - x^3) $$

1.5 Differentiate the following:

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