Part IV Calculate the following three limits ( 5 marks $$ imes 3=15 $$ marks). 16. $$ \lim _{x ightarrow \pi} \frac{2 x^{2}-6 \pi x+4 \pi^{2}}{x^{2}-\pi^{2}} $$ 17. $$ \lim _{x ightarrow 6^{+}} \frac{4(x-6)}{\sqrt{x^{2}-36}} $$ 18. $$ \lim _{t ightarrow 0} \frac{1-\cos t}{\sin t an t} $$ Part V Find the first derivative $$ y^{\prime} $$ of the following three functions ( 5 marks $$ x $$ $$ 3=15 $$ marks). 19. $$ y=\frac{3 x^{2}-2}{x+1} $$ 20. $$ y=\sin \left(x^{2}-3 x ight)-3 x^{2} \cos x $$ 21. $$ \sqrt{y}+x y^{2}=5 $$ Page 2 of 3

Question

Part IV Calculate the following three limits ( 5 marks $$ 	imes 3=15 $$ marks). 16. $$ \lim _{x ightarrow \pi} \frac{2 x^{2}-6 \pi x+4 \pi^{2}}{x^{2}-\pi^{2}} $$ 17. $$ \lim _{x ightarrow 6^{+}} \frac{4(x-6)}{\sqrt{x^{2}-36}} $$ 18. $$ \lim _{t ightarrow 0} \frac{1-\cos t}{\sin t 	an t} $$ Part V Find the first derivative $$ y^{\prime} $$ of the following three functions ( 5 marks $$ x $$ $$ 3=15 $$ marks). 19. $$ y=\frac{3 x^{2}-2}{x+1} $$ 20. $$ y=\sin \left(x^{2}-3 xight)-3 x^{2} \cos x $$ 21. $$ \sqrt{y}+x y^{2}=5 $$ Page 2 of 3

UpStudy AI · Accepted Answer

Sure, let's tackle each of these problems step by step.

---

## **Part IV: Calculate the Limits**

### **16.**
$$
\lim _{x \rightarrow \pi} \frac{2 x^{2} - 6 \pi x + 4 \pi^{2}}{x^{2} - \pi^{2}}
$$

**Solution:**

First, substitute $$ x = \pi $$ to check for indeterminate forms:

$$
\frac{2\pi^{2} - 6\pi^{2} + 4\pi^{2}}{\pi^{2} - \pi^{2}} = \frac{0}{0}
$$

Since it's an indeterminate form $$ \frac{0}{0} $$, we can factor both the numerator and the denominator.

**Factor the numerator:**

$$
2x^{2} - 6\pi x + 4\pi^{2} = 2(x^{2} - 3\pi x + 2\pi^{2}) = 2(x - \pi)(x - 2\pi)
$$

**Factor the denominator:**

$$
x^{2} - \pi^{2} = (x - \pi)(x + \pi)
$$

Now, the expression becomes:

$$
\frac{2(x - \pi)(x - 2\pi)}{(x - \pi)(x + \pi)} = \frac{2(x - 2\pi)}{x + \pi} \quad \text{(for } x \neq \pi \text{)}
$$

Now, take the limit as $$ x \rightarrow \pi $$:

$$
\frac{2(\pi - 2\pi)}{\pi + \pi} = \frac{2(-\pi)}{2\pi} = -1
$$

**Answer:** $$-1$$

---

### **17.**
$$
\lim _{x \rightarrow 6^{+}} \frac{4(x - 6)}{\sqrt{x^{2} - 36}}
$$

**Solution:**

First, note that as $$ x \rightarrow 6^{+} $$, $$ x > 6 $$, so $$ \sqrt{x^{2} - 36} = \sqrt{(x - 6)(x + 6)} $$.

Simplify the expression:

$$
\frac{4(x - 6)}{\sqrt{(x - 6)(x + 6)}} = \frac{4(x - 6)}{\sqrt{x - 6} \cdot \sqrt{x + 6}} = \frac{4\sqrt{x - 6}}{\sqrt{x + 6}}
$$

Now, as $$ x \rightarrow 6^{+} $$:
- $$ x - 6 \rightarrow 0^{+} $$, so $$ \sqrt{x - 6} \rightarrow 0 $$
- $$ \sqrt{x + 6} \rightarrow \sqrt{12} = 2\sqrt{3} $$

Thus,

$$
\lim _{x \rightarrow 6^{+}} \frac{4\sqrt{x - 6}}{2\sqrt{3}} = \frac{4 \cdot 0}{2\sqrt{3}} = 0
$$

**Answer:** $$0$$

---

### **18.**
$$
\lim _{t \rightarrow 0} \frac{1 - \cos t}{\sin t \tan t}
$$

**Solution:**

First, recall that $$ \tan t = \frac{\sin t}{\cos t} $$. Substitute this into the expression:

$$
\frac{1 - \cos t}{\sin t \cdot \frac{\sin t}{\cos t}} = \frac{1 - \cos t}{\frac{\sin^{2} t}{\cos t}} = \frac{(1 - \cos t) \cos t}{\sin^{2} t}
$$

Now, use the trigonometric identity $$ 1 - \cos t = 2 \sin^{2} \left(\frac{t}{2}\right) $$:

$$
\frac{2 \sin^{2} \left(\frac{t}{2}\right) \cos t}{\sin^{2} t} = \frac{2 \sin^{2} \left(\frac{t}{2}\right) \cos t}{4 \sin^{2} \left(\frac{t}{2}\right) \cos^{2} \left(\frac{t}{2}\right)}}
$$

Simplify:

$$
\frac{2 \cos t}{4 \cos^{2} \left(\frac{t}{2}\right)} = \frac{\cos t}{2 \cos^{2} \left(\frac{t}{2}\right)}
$$

Using the double-angle identity $$ \cos t = 2 \cos^{2} \left(\frac{t}{2}\right) - 1 $$, but it's easier to recognize that as $$ t \rightarrow 0 $$:

$$
\lim _{t \rightarrow 0} \frac{1 - \cos t}{\sin t \tan t} = \lim _{t \rightarrow 0} \frac{1 - \cos t}{\sin^{2} t / \cos t} = \lim _{t \rightarrow 0} \frac{(1 - \cos t) \cos t}{\sin^{2} t}
$$

Using the standard limit $$ \lim_{t \to 0} \frac{1 - \cos t}{t^{2}} = \frac{1}{2} $$, and $$ \lim_{t \to 0} \frac{\cos t}{\sin^{2} t / t^{2}} = \infty $$, but here the correct approach is to use L’Hôpital’s Rule:

Original limit:
$$
\frac{1 - \cos t}{\sin t \tan t} = \frac{1 - \cos t}{\sin^{2} t / \cos t} = \frac{(1 - \cos t) \cos t}{\sin^{2} t}
$$

Differentiate numerator and denominator:

Numerator derivative:
$$
\frac{d}{dt}[(1 - \cos t)\cos t] = \sin t \cos t + (1 - \cos t)(-\sin t) = \sin t \cos t - \sin t + \sin t \cos t = 2 \sin t \cos t - \sin t
$$

Denominator derivative:
$$
\frac{d}{dt}[\sin^{2} t] = 2 \sin t \cos t
$$

Now, the limit becomes:

$$
\lim_{t \to 0} \frac{2 \sin t \cos t - \sin t}{2 \sin t \cos t} = \lim_{t \to 0} \frac{\sin t (2 \cos t - 1)}{2 \sin t \cos t} = \lim_{t \to 0} \frac{2 \cos t - 1}{2 \cos t}
$$

Now, substitute $$ t = 0 $$:

$$
\frac{2(1) - 1}{2(1)} = \frac{1}{2}
$$

**Answer:** $$\dfrac{1}{2}$$

---

## **Part V: Find the First Derivative $$ y' $$**

### **19.**
$$
y = \frac{3x^{2} - 2}{x + 1}
$$

**Solution:**

Use the **Quotient Rule**:
$$
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^{2}}
$$

Let $$ u = 3x^{2} - 2 $$ and $$ v = x + 1 $$.

Compute derivatives:
$$
u' = 6x \quad \text{and} \quad v' = 1
$$

Apply the Quotient Rule:
$$
y' = \frac{(x + 1)(6x) - (3x^{2} - 2)(1)}{(x + 1)^{2}} = \frac{6x(x + 1) - (3x^{2} - 2)}{(x + 1)^{2}}
$$

Expand and simplify the numerator:
$$
6x^{2} + 6x - 3x^{2} + 2 = 3x^{2} + 6x + 2
$$

Thus,
$$
y' = \frac{3x^{2} + 6x + 2}{(x + 1)^{2}}
$$

**Answer:** $$ y' = \dfrac{3x^{2} + 6x + 2}{(x + 1)^{2}} $$

---

### **20.**
$$
y = \sin(x^{2} - 3x) - 3x^{2} \cos x
$$

**Solution:**

Differentiate each term separately.

**First term:** $$ \sin(x^{2} - 3x) $$

Use the **Chain Rule**:
$$
\frac{d}{dx} \sin(u) = \cos(u) \cdot u'
$$
where $$ u = x^{2} - 3x $$, so $$ u' = 2x - 3 $$.

Thus,
$$
\frac{d}{dx} \sin(x^{2} - 3x) = \cos(x^{2} - 3x) \cdot (2x - 3)
$$

**Second term:** $$ -3x^{2} \cos x $$

Use the **Product Rule**:
$$
\frac{d}{dx} [u \cdot v] = u' \cdot v + u \cdot v'
$$
where $$ u = -3x^{2} $$ and $$ v = \cos x $$.

Compute derivatives:
$$
u' = -6x \quad \text{and} \quad v' = -\sin x
$$

Thus,
$$
\frac{d}{dx} (-3x^{2} \cos x) = (-6x) \cos x + (-3x^{2})(- \sin x) = -6x \cos x + 3x^{2} \sin x
$$

**Combine both derivatives:**
$$
y' = \cos(x^{2} - 3x)(2x - 3) - 6x \cos x + 3x^{2} \sin x
$$

**Answer:**
$$
y' = (2x - 3)\cos(x^{2} - 3x) - 6x\cos x + 3x^{2}\sin x
$$

---

### **21.**
$$
\sqrt{y} + x y^{2} = 5
$$

**Solution:**

We need to find $$ \frac{dy}{dx} $$ (denoted as $$ y' $$). Use **Implicit Differentiation**.

Differentiate both sides with respect to $$ x $$:

$$
\frac{d}{dx} \left( \sqrt{y} \right) + \frac{d}{dx} \left( x y^{2} \right) = \frac{d}{dx} (5)
$$

Compute each derivative:

1. **First term:** $$ \frac{d}{dx} \sqrt{y} = \frac{1}{2\sqrt{y}} \cdot y' $$

2. **Second term:** Use the **Product Rule** for $$ x y^{2} $$:
$$
\frac{d}{dx} (x y^{2}) = y^{2} \cdot \frac{d}{dx} (x) + x \cdot \frac{d}{dx} (y^{2}) = y^{2} + x \cdot 2y y' = y^{2} + 2x y y'
$$

3. **Right side:** $$ \frac{d}{dx} (5) = 0 $$

Now, combine all:

$$
\frac{1}{2\sqrt{y}} y' + y^{2} + 2x y y' = 0
$$

Factor out $$ y' $$:

$$
y' \left( \frac{1}{2\sqrt{y}} + 2x y \right) = -y^{2}
$$

Solve for $$ y' $$:

$$
y' = \frac{ -y^{2} }{ \frac{1}{2\sqrt{y}} + 2x y } = \frac{ -y^{2} }{ \frac{1 + 4x y \sqrt{y} }{2\sqrt{y}} } = \frac{ -2 y^{2} \sqrt{y} }{ 1 + 4x y \sqrt{y} }
$$

Simplify:

$$
y' = \frac{ -2 y^{2.5} }{ 1 + 4x y^{1.5} }
$$

Alternatively, keeping it in terms of radicals:

$$
y' = \frac{ -2 y^{2} \sqrt{y} }{ 1 + 4x y \sqrt{y} } = \frac{ -2 y^{5/2} }{ 1 + 4x y^{3/2} }
$$

**Answer:**
$$
y' = \dfrac{ -2 y^{5/2} }{ 1 + 4x y^{3/2} }
$$

---

## **Summary of Answers**

16. $$-1$$

17. $$0$$

18. $$\dfrac{1}{2}$$

19. $$ y' = \dfrac{3x^{2} + 6x + 2}{(x + 1)^{2}} $$

20. $$ y' = (2x - 3)\cos(x^{2} - 3x) - 6x\cos x + 3x^{2}\sin x $$

21. $$ y' = \dfrac{ -2 y^{5/2} }{ 1 + 4x y^{3/2} } $$
**Part IV: Limits** 16. $$-1$$ 17. $$0$$ 18. $$\dfrac{1}{2}$$ **Part V: Derivatives** 19. $$ y' = \dfrac{3x^{2} + 6x + 2}{(x + 1)^{2}} $$ 20. $$ y' = (2x - 3)\cos(x^{2} - 3x) - 6x\cos x + 3x^{2}\sin x $$ 21. $$ y' = \dfrac{ -2 y^{5/2} }{ 1 + 4x y^{3/2} } $$

Part IV Calculate the following three limits ( 5 marks marks).
16.
17.
18.
Part V Find the first derivative of the following three functions ( 5 marks
marks).
19.
20.
21.
Page 2 of 3

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Calculus Questions

Part IV Calculate the following three limits ( 5 marks marks). 16. 17. 18. Part V Find the first derivative of the following three functions ( 5 marks marks). 19. 20. 21. Page 2 of 3

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Calculus Questions

Part IV Calculate the following three limits ( 5 marks marks).
16.
17.
18.
Part V Find the first derivative of the following three functions ( 5 marks
marks).
19.
20.
21.
Page 2 of 3