3 Differentiate each of the following with respect to $$ x $$ (a) $$ \sin ^{5} x-9 \sin x $$ (b) $$ \cos ^{4}\left(4 x^{4}\right) $$ (c) $$ \cos \sqrt{x^{2}+7} $$ (d) $$ \frac{1}{\sqrt{\sin ^{2} 3 x+7}} $$ (e) $$ (2-\cos 2 x)^{3} $$ (f) $$ \sqrt{3-\sin 2 x} $$

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**(a)**  
Differentiate  
$$
y = \sin^5 x - 9 \sin x.
$$  
1. For $$\sin^5 x$$:  
  Differentiate using the chain rule:  
  $$
  \frac{d}{dx}\left(\sin^5 x\right)=5\,\sin^4 x\cdot\cos x.
  $$  
2. For $$-9\sin x$$:  
  $$
  \frac{d}{dx}\left(-9\sin x\right)=-9\cos x.
  $$  
Thus, combining the results:  
$$
\frac{dy}{dx}=5\,\sin^4 x\,\cos x-9\cos x=\cos x\Bigl(5\,\sin^4 x-9\Bigr).
$$

---

**(b)**  
Differentiate  
$$
y=\cos^4(4x^4).
$$  
1. Let $$u=\cos(4x^4)$$ so that $$y=u^4$$. Then,  
  $$\frac{dy}{du}=4u^3$$.  
2. Next, differentiate $$u=\cos(4x^4)$$:  
  $$\frac{du}{dx}=-\sin(4x^4)\cdot\frac{d}{dx}(4x^4)$$.  
  Since $$\frac{d}{dx}(4x^4)=16x^3$$, we have  
  $$
  \frac{du}{dx}=-16x^3\sin(4x^4).
  $$  
3. By the chain rule:  
$$
\frac{dy}{dx}=4\,[\cos(4x^4)]^3\cdot\Bigl(-16x^3\sin(4x^4)\Bigr)
=-64x^3\,\cos^3(4x^4)\,\sin(4x^4).
$$

---

**(c)**  
Differentiate  
$$
y=\cos\sqrt{x^2+7}.
$$  
1. Let $$v=\sqrt{x^2+7}=(x^2+7)^{1/2}$$. Then,  
  $$\frac{dv}{dx}=\frac{1}{2}(x^2+7)^{-1/2}\cdot2x=\frac{x}{\sqrt{x^2+7}}$$.  
2. Differentiate $$y=\cos v$$:  
  $$\frac{dy}{dv}=-\sin v$$.  
3. Applying the chain rule:  
$$
\frac{dy}{dx}=-\sin\bigl(\sqrt{x^2+7}\bigr)\cdot\frac{x}{\sqrt{x^2+7}}
=-\frac{x\,\sin\bigl(\sqrt{x^2+7}\bigr)}{\sqrt{x^2+7}}.
$$

---

**(d)**  
Differentiate  
$$
y=\frac{1}{\sqrt{\sin^2 3x+7}}.
$$  
Write the function as  
$$
y=(\sin^2 3x+7)^{-1/2}.
$$  
1. Differentiating with respect to $$u=\sin^2 3x+7$$:  
  $$\frac{dy}{du}=-\frac{1}{2}u^{-3/2}$$.  
2. Now, differentiate $$u=\sin^2 3x+7$$.  
  For $$\sin^2 3x$$, let $$w=\sin 3x$$ so that $$w^2$$ differentiates as  
  $$\frac{d}{dx}(w^2)=2w\cdot\frac{dw}{dx}$$.  
  Since $$\frac{d}{dx}(\sin 3x)=3\cos3x$$, we obtain  
  $$
  \frac{d}{dx}(\sin^2 3x)=2\sin3x\cdot3\cos3x=6\sin3x\cos3x.
  $$
3. Thus, by the chain rule:  
$$
\frac{dy}{dx}=-\frac{1}{2}(\sin^2 3x+7)^{-3/2}\cdot6\sin3x\cos3x
=-3\sin3x\cos3x\,(\sin^2 3x+7)^{-3/2}.
$$

---

**(e)**  
Differentiate  
$$
y=(2-\cos2x)^3.
$$  
1. Let $$u=2-\cos2x$$ so that $$y=u^3$$ and $$\frac{dy}{du}=3u^2$$.  
2. Differentiate $$u=2-\cos2x$$:  
  $$\frac{du}{dx}=0-(-\sin2x\cdot2)=2\sin2x$$.  
3. Thus, by the chain rule:  
$$
\frac{dy}{dx}=3(2-\cos2x)^2\cdot2\sin2x=6(2-\cos2x)^2\sin2x.
$$

---

**(f)**  
Differentiate  
$$
y=\sqrt{3-\sin2x}=(3-\sin2x)^{1/2}.
$$  
1. Differentiate the outer function:  
  $$\frac{dy}{du}=\frac{1}{2}(3-\sin2x)^{-1/2}$$ where $$u=3-\sin2x$$.  
2. Differentiate $$u=3-\sin2x$$:  
  $$\frac{du}{dx}=-\cos2x\cdot2=-2\cos2x$$.  
3. Thus, by the chain rule:  
$$
\frac{dy}{dx}=\frac{1}{2}(3-\sin2x)^{-1/2}\cdot(-2\cos2x)
=-\cos2x\,(3-\sin2x)^{-1/2}=-\frac{\cos2x}{\sqrt{3-\sin2x}}.
$$
**(a)** $$ \frac{dy}{dx} = \cos x \left(5 \sin^4 x - 9\right) $$ **(b)** $$ \frac{dy}{dx} = -64x^3 \cos^3(4x^4) \sin(4x^4) $$ **(c)** $$ \frac{dy}{dx} = -\frac{x \sin\left(\sqrt{x^2 + 7}\right)}{\sqrt{x^2 + 7}} $$ **(d)** $$ \frac{dy}{dx} = -3 \sin 3x \cos 3x \left(\sin^2 3x + 7\right)^{-3/2} $$ **(e)** $$ \frac{dy}{dx} = 6 (2 - \cos 2x)^2 \sin 2x $$ **(f)** $$ \frac{dy}{dx} = -\frac{\cos 2x}{\sqrt{3 - \sin 2x}} $$

3 Differentiate each of the following with
respect to
(a)
(b)
©
(d)
(e)
(f)

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3 Differentiate each of the following with respect to (a) (b) © (d) (e) (f)