QE 4.4. (1) Find dy (a) y = 4x3 − 2x + 5. (b) y = 2sqrt(x4 + 6x). (c) y = cos(x3 − 5x + 11). (d) y = (x10 + sin(2x)).

$$dy = \left(12x^2 - 2\right) \, dx$$

The differential is $\boxed{dy = (12x^2 - 2) \, dx}$.

$$\frac{dy}{dx} = 2 \cdot \frac{1}{2}(x^4 + 6x)^{(1/2)-1} \cdot \frac{d}{dx}(x^4 + 6x)$$ $$\frac{dy}{dx} = (x^4 + 6x)^{-1/2} \cdot (4x^3 + 6)$$ $$\frac{dy}{dx} = \frac{4x^3 + 6}{\sqrt{x^4 + 6x}}$$

$$dy = \frac{4x^3 + 6}{\sqrt{x^4 + 6x}} \, dx$$

The differential is $\boxed{dy = \frac{4x^3 + 6}{\sqrt{x^4 + 6x}} \, dx}$.

Let $u = x^3 - 5x + 11$. Then $y = \cos u$.

$$\frac{dy}{dx} = \frac{d}{du}(\cos u) \cdot \frac{du}{dx}$$ $$\frac{dy}{dx} = (-\sin u) \cdot \frac{d}{dx}(x^3 - 5x + 11)$$ $$\frac{dy}{dx} = -\sin(x^3 - 5x + 11) \cdot (3x^2 - 5)$$ $$\frac{dy}{dx} = -(3x^2 - 5)\sin(x^3 - 5x + 11)$$

$$dy = -(3x^2 - 5)\sin(x^3 - 5x + 11) \, dx$$

The differential is $\boxed{dy = -(3x^2 - 5)\sin(x^3 - 5x + 11) \, dx}$.

$$\frac{dy}{dx} = \frac{d}{dx}(x^{10}) + \frac{d}{dx}(\sin(2x))$$

For $\frac{d}{dx}(\sin(2x))$, use the chain rule. Let $u = 2x$, then $\frac{d}{dx}(\sin u) = \cos u \cdot \frac{du}{dx} = \cos(2x) \cdot 2 = 2\cos(2x)$.

$$\frac{dy}{dx} = 10x^9 + 2\cos(2x)$$

$$dy = (10x^9 + 2\cos(2x)) \, dx$$

The differential is $\boxed{dy = (10x^9 + 2\cos(2x)) \, dx}$.