Test 2 Review Material

Below are resources to help you practice and prepare for the test 2 in this course.

Test 2 will cover the material covered in Chapter 4 of this textbook:

  • basic derivative rules
  • product, quotient and chain rules
  • implicit derivatives

What’s not on this test:

  • no interpretation style questions on this test
  • no derivations
  • no limit definitions

Test Instructions

Below are the general instructions for all tests

  • You will get a chance to retake this. Your highest score counts toward your final grade.
  • Follow the guidance for each part and show all work for full credit.
  • Non-graphical calculators are allowed.
  • One page (two sides) of handwritten (or font size 8) notes are allowed.
  • The equation sheet from the textbook is included as part of the test pack (it doesn’t count as note pages)

Basic Rule Practice

1. \(f(x)=7\)

Math Hint The derivative of any constant is always \(0\).
Think of a flat line — its slope never changes.
Solution \(f'(x)=0\)

2. \(f(x)=x^5\)

Math Hint Use the power rule: \((x^n)' = n x^{n-1}\).
Solution \(f'(x)=5x^4\)

3. \(f(x)=3x^4\)

Math Hint You have a constant multiplied by a power of \(x\).
Apply the constant multiple rule with the power rule.
Solution \(f'(x)=12x^3\)

4. \(f(x)=5x^3-2x^2\)

Math Hint Use the difference rule: differentiate each term individually.
Solution \(f'(x)=15x^2-4x\)

5. \(f(x)=x^4+3x^2+5\)

Math Hint Use the sum rule and note that the derivative of a constant is zero.
Solution \(f'(x)=4x^3+6x\)

6. \(f(x)=9x^7-4x^3+2x\)

Math Hint Each term loses one power when you apply the power rule.
Solution \(f'(x)=63x^6-12x^2+2\)

7. \(f(x)=\dfrac{1}{2}x^2-3x+4\)

Math Hint Differentiate each term.
The first term is a parabola — its slope changes linearly with \(x\).
Solution \(f'(x)=x-3\)

8. \(f(x)=10x^3+7x-12\)

Math Hint Differentiate each term using the power rule.
Solution \(f'(x)=30x^2+7\)

9. \(f(x)=x^6-9x^2+1\)

Math Hint Find \(f'(x)\) and simplify.
Solution \(f'(x)=6x^5-18x=6x(x^4-3)\)

10. \(f(x)=4x^3-12x^2+9x-1\)

Math Hint Differentiate each term; the derivative of a cubic is a quadratic.
Solution \(f'(x)=12x^2-24x+9\)

Identity Rule Practice

1. \(f(x)=e^x\)

Math Hint Use the table entry for \(e^x\).
Solution \(f'(x)=e^x\)

2. \(f(x)=3e^x\)

Math Hint Differentiate \(e^x\) from the table, then scale by the constant.
Solution \(f'(x)=3e^x\)

3. \(f(x)=2e^{2x}\)

Math Hint Use the identity for \(e^{g(x)}\) with \(g(x)=2x\).
Solution \(f'(x)=4e^{2x}\)

4. \(f(x)=5e^{-x}\)

Math Hint Use the identity for \(e^{g(x)}\) with \(g(x)=-x\).
Solution \(f'(x)=-5e^{-x}\)

5. \(f(x)=e^{x^2}\)

Math Hint Use the identity for \(e^{g(x)}\) with \(g(x)=x^2\).
Solution \(f'(x)=2x e^{x^2}\)

6. \(f(x)=e^{3x+1}\)

Math Hint Use the identity for \(e^{g(x)}\) with \(g(x)=3x+1\).
Solution \(f'(x)=3e^{3x+1}\)

7. \(f(x)=10^x\)

Math Hint Use the table entry for \(a^x\) with \(a=10\).
Solution \(f'(x)=10^x \ln(10)\)

8. \(f(x)=2^{3x}\)

Math Hint Use the identity for \(a^{g(x)}\) with \(a=2\), \(g(x)=3x\).
Solution \(f'(x)=3\cdot 2^{3x}\ln 2\)

9. \(f(x)=5^x + 7^x\)

Math Hint Apply the table entry for each exponential term separately.
Solution \(f'(x)=5^x\ln(5)+7^x\ln(7)\)

10. \(f(x)=3^x - 2^x\)

Math Hint Use the \(a^x\) identity on each term.
Solution \(f'(x)=3^x\ln(3)-2^x\ln(2)\)

11. \(f(x)=\ln(x)\)

Math Hint Use the table entry for \(\ln(x)\) (domain \(x>0\)).
Solution \(f'(x)=\dfrac{1}{x}\)

12. \(f(x)=\ln(3x)\)

Math Hint Use the identity for \(\ln(g(x))\) with \(g(x)=3x\) (domain \(x>0\)).
Solution \(f'(x)=\dfrac{1}{x}\)

13. \(f(x)=\ln(x^2+1)\)

Math Hint Use the identity for \(\ln(g(x))\) with \(g(x)=x^2+1\).
Solution \(f'(x)=\dfrac{2x}{x^2+1}\)

14. \(f(x)=\ln(\sqrt{x})\)

Math Hint Write \(\sqrt{x}=x^{1/2}\), then use the \(\ln(x)\) identity (domain \(x>0\)).
Solution \(f'(x)=\dfrac{1}{2x}\)

15. \(f(x)=\log_2(x)\)

Math Hint Use the table entry for \(\log_a(x)\) with \(a=2\) (domain \(x>0\)).
Solution \(f'(x)=\dfrac{1}{x\ln(2)}\)

16. \(f(x)=\log_3(2x)\)

Math Hint Use the identity for \(\log_a(g(x))\) with \(a=3\), \(g(x)=2x\) (domain \(x>0\)).
Solution \(f'(x)=\dfrac{1}{x\ln(3)}\)

17. \(f(x)=\ln(e^x+1)\)

Math Hint Use the identity for \(\ln(g(x))\) with \(g(x)=e^x+1\).
Solution \(f'(x)=\dfrac{e^x}{e^x+1}\)

18. \(f(x)=e^{\ln(x)}\)

Math Hint First simplify using \(e^{\ln(x)}=x\) (domain \(x>0\)).
Solution \(f'(x)=1\)

19. \(f(x)=x e^x\)

Math Hint Differentiate each factor using the table entries \(d(x)/dx=1\) and \(d(e^x)/dx=e^x\), then combine appropriately.
Solution \(f'(x)=e^x(1+x)\)

20. \(f(x)=\ln(x^3+x)\)

Math Hint Use the identity for \(\ln(g(x))\) with \(g(x)=x^3+x\) (domain \(x>0\)).
Solution \(f'(x)=\dfrac{3x^2+1}{x^3+x}\)

Basic Trig Derivative Practice

1. \(f(x)=\sin x\)

Math Hint Use \((\sin x)'=\cos x\).
Solution \(f'(x)=\cos x\)

2. \(f(x)=\cos x\)

Math Hint \((\cos x)'=-\sin x\)
Solution \(f'(x)=-\sin x\)

3. \(f(x)=2\sin x\)

Math Hint Multiply the derivative of \(\sin x\) by 2.
Solution \(f'(x)=2\cos x\)

4. \(f(x)=5\cos x\)

Math Hint \((\cos x)'=-\sin x\)
Solution \(f'(x)=-5\sin x\)

5. \(f(x)=\sin x+\cos x\)

Math Hint Differentiate each term separately.
Solution \(f'(x)=\cos x-\sin x\)

6. \(f(x)=3\sin x-4\cos x\)

Math Hint Use constant multiple and sum/difference rules.
Solution \(f'(x)=3\cos x+4\sin x\)

7. \(f(x)=\tan x\)

Math Hint \((\tan x)'=\sec^2 x\)
Solution \(f'(x)=\sec^2 x\)

8. \(f(x)=\sec x\)

Math Hint \((\sec x)'=\sec x\tan x\)
Solution \(f'(x)=\sec x\tan x\)

9. \(f(x)=\cot x\)

Math Hint \((\cot x)'=-\csc^2 x\)
Solution \(f'(x)=-\csc^2 x\)

10. \(f(x)=\csc x\)

Math Hint \((\csc x)'=-\csc x\cot x\)
Solution \(f'(x)=-\csc x\cot x\)

Product Rule Practice

1. \(f(x)=x\sin x\)

Math Hint Use the product rule: \((uv)' = u'v + uv'\)
Solution \(f'(x)=\sin x + x\cos x\)

2. \(f(x)=x\cos x\)

Math Hint \((\cos x)'=-\sin x\)
Solution \(f'(x)=\cos x - x\sin x\)

3. \(f(x)=x^2 e^x\)

Math Hint Apply product rule: \(u=x^2, v=e^x\)
Solution \(f'(x)=2x e^x + x^2 e^x = e^x(x^2 + 2x)\)

4. \(f(x)=x^3\ln x\)

Math Hint Use the product rule and derivative of \(\ln x\).
Solution \(f'(x)=3x^2\ln x + x^2\)

5. \(f(x)=(x^2+1)\sin x\)

Math Hint Differentiate both terms using product rule.
Solution \(f'(x)=2x\sin x + (x^2+1)\cos x\)

6. \(f(x)=x e^{2x}\)

Math Hint Product rule with \(v' = 2e^{2x}\)
Solution \(f'(x)=e^{2x}(1+2x)\)

7. \(f(x)=x^2\cos x\)

Math Hint Product rule: \(u=x^2, v=\cos x\)
Solution \(f'(x)=2x\cos x - x^2\sin x\)

8. \(f(x)=(3x-2)(x^2+4)\)

Math Hint Use product rule or expand first.
Solution \(f'(x)=9x^2-4x+12\)

9. \(f(x)=(x+1)e^x\)

Math Hint Use product rule and factor \(e^x\).
Solution \(f'(x)=e^x(x+2)\)

10. \(f(x)=(2x-5)\tan x\)

Math Hint Use product rule: \(u=2x-5, v=\tan x\)
Solution \(f'(x)=2\tan x + (2x-5)\sec^2 x\)

Quotient Rule Practice

1. \(f(x)=\dfrac{1}{x}\)

Math Hint Rewrite as \(x^{-1}\) or use the quotient rule.
Solution \(f'(x)=-\dfrac{1}{x^2}\)

2. \(f(x)=\dfrac{x}{x^2+1}\)

Math Hint Use \((u/v)'=\dfrac{u'v-uv'}{v^2}\)
Solution \(f'(x)=\dfrac{1-x^2}{(x^2+1)^2}\)

3. \(f(x)=\dfrac{x^2}{x+1}\)

Math Hint Apply quotient rule and simplify.
Solution \(f'(x)=\dfrac{x^2+2x}{(x+1)^2}\)

4. \(f(x)=\dfrac{x^2+1}{x}\)

Math Hint Simplify after differentiating.
Solution \(f'(x)=1-\dfrac{1}{x^2}\)

5. \(f(x)=\dfrac{\sin x}{x}\)

Math Hint Use quotient rule: \(u=\sin x, v=x\)
Solution \(f'(x)=\dfrac{x\cos x-\sin x}{x^2}\)

6. \(f(x)=\dfrac{x}{\sin x}\)

Math Hint Differentiate and simplify.
Solution \(f'(x)=\dfrac{\sin x - x\cos x}{\sin^2 x}\)

7. \(f(x)=\dfrac{e^x}{x}\)

Math Hint Combine exponential and quotient rules.
Solution \(f'(x)=\dfrac{e^x(x-1)}{x^2}\)

8. \(f(x)=\dfrac{\ln x}{x}\)

Math Hint Use quotient rule: \(u=\ln x, v=x\)
Solution \(f'(x)=\dfrac{1-\ln x}{x^2}\)

9. \(f(x)=\dfrac{x^2}{e^x}\)

Math Hint Differentiate numerator and denominator.
Solution \(f'(x)=\dfrac{2x-x^2}{e^x}\)

10. \(f(x)=\dfrac{\tan x}{x^2}\)

Math Hint Use quotient rule and simplify.
Solution \(f'(x)=\dfrac{x\sec^2 x - 2\tan x}{x^3}\)

Chain Rule Practice

1. \(f(x)=\sin(2x)\)

Math Hint Outer: \(\sin(u)\), Inner: \(u=2x\)
Solution \(f'(x)=2\cos(2x)\)

2. \(f(x)=\cos(3x)\)

Math Hint Outer: \(\cos(u)\), Inner: \(u=3x\)
Solution \(f'(x)=-3\sin(3x)\)

3. \(f(x)=e^{5x}\)

Math Hint Use \((e^{g(x)})' = g'(x)e^{g(x)}\)
Solution \(f'(x)=5e^{5x}\)

4. \(f(x)=\ln(4x+1)\)

Math Hint Use chain rule for \(\ln(g(x))\)
Solution \(f'(x)=\dfrac{4}{4x+1}\)

5. \(f(x)=\sqrt{3x+2}\)

Math Hint \(\sqrt{3x+2}=(3x+2)^{1/2}\)
Solution \(f'(x)=\dfrac{3}{2\sqrt{3x+2}}\)

6. \(f(x)=(2x+1)^5\)

Math Hint Use \((g(x))^n)'=n(g(x))^{n-1}g'(x)\)
Solution \(f'(x)=10(2x+1)^4\)

7. \(f(x)=\tan(4x)\)

Math Hint Outer: \(\tan(u)\), Inner: \(u=4x\)
Solution \(f'(x)=4\sec^2(4x)\)

8. \(f(x)=\sin(x^2)\)

Math Hint Outer: \(\sin(u)\), Inner: \(u=x^2\)
Solution \(f'(x)=2x\cos(x^2)\)

9. \(f(x)=e^{x^2+1}\)

Math Hint Differentiate exponent using chain rule.
Solution \(f'(x)=2x e^{x^2+1}\)

10. \(f(x)=\cos(\sqrt{x})\)

Math Hint Outer: \(\cos(u)\); Inner: \(u=\sqrt{x}\)
Solution \(f'(x)=-\dfrac{\sin(\sqrt{x})}{2\sqrt{x}}\)

Implicit Derivative Practice

1. \(x^2 + y^2 = 25\)

Math Hint Differentiate both sides with respect to \(x\).
Solution \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)

2. \(x^3 + y^3 = 9xy\)

Math Hint Use the product rule on \(9xy\).
Solution \(\dfrac{dy}{dx} = \dfrac{9y - 3x^2}{3y^2 - 9x}\)

3. \(x^2y + xy^2 = 6\)

Math Hint Apply product rule to both terms.
Solution \(\dfrac{dy}{dx} = -\dfrac{2xy + y^2}{x^2 + 2xy}\)

4. \(\sin y = x^2 + y\)

Math Hint Differentiate \(\sin y\) using chain rule.
Solution \(\dfrac{dy}{dx} = \dfrac{2x}{\cos y - 1}\)

5. \(e^y = x^2y + 3\)

Math Hint Differentiate each term, use product rule for \(x^2y\).
Solution \(\dfrac{dy}{dx} = \dfrac{2xy}{e^y - x^2}\)

6. \(\tan y = x\)

Math Hint Differentiate \(\tan y\) with chain rule.
Solution \(\dfrac{dy}{dx}=\cos^2 y\)

7. \(x^2 + xy + y^2 = 7\)

Math Hint Use product rule on \(xy\) and chain rule on \(y^2\).
Solution \(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}\)

8. \(x^2 = \sin y\)

Math Hint Differentiate both sides.
Solution \(\dfrac{dy}{dx}=\dfrac{2x}{\cos y}\)

9. \(x^2y^2 = 16\)

Math Hint Product rule: \(u=x^2, v=y^2\)
Solution \(\dfrac{dy}{dx} = -\dfrac{y}{x}\)

10. \(x^2 + \cos y = 4\)

Math Hint Derivative of \(\cos y\) is \(-\sin y\dfrac{dy}{dx}\).
Solution \(\dfrac{dy}{dx} = \dfrac{2x}{\sin y}\)

Test 2 — Practice Test I

How it works. Each problem has a locked, step-by-step solution.
To unlock, translate the given Australian slang into US English.
Type the translation as lowercase with no spaces/punctuation.
(Example: “gas station” → gasstation)

NOTE: The actual test will NOT include follow-ups — they are for review.

SOLUTIONS: may have errors - please reach out if something doesn’t seem right

🔒 Q1

Problem. Find \(f'(x)\) for \(F(x)=4x^3-5x^2+6x-2\).

Follow-up. At what \(x\)-value is the slope zero? What does that tell you about the graph?

Click to Unlock Solution — Password prompt: Translate arvo → US English

🔒 Q2

Problem. Differentiate \(F(x)=\dfrac{4e^{3x}}{x^2}=4e^{3x}x^{-2}\).

Follow-up. As \(x\) increases, does exponential growth or division by \(x^2\) dominate the slope?

Click to Unlock Solution — Password prompt: Translate sunnies → US English

🔒 Q3

Problem. Find \(F'(x)\) for \(F(x)=e^x + x e^2 + 4e\).

Click to Unlock Solution — Password prompt: Translate mozzie → US English

🔒 Q4

Problem. Find \(F'(x)\) for \(F(x)=x^2 e^{x^3}\).

Click to Unlock Solution — Password prompt: Translate servo → US English

🔒 Q5

Problem. Differentiate \(F(x)=\dfrac{\ln x}{x^3}\) for \(x>0\).

Click to Unlock Solution — Password prompt: Translate brekkie → US English

🔒 Q6

Problem. Find \(F'(x)\) for \(F(x)=\ln(2x+5)\).

Click to Unlock Solution — Password prompt: Translate thongs (footwear) → US English

🔒 Q7

Problem. Differentiate \(F(x)=e^{x^2}\sin x\).

Click to Unlock Solution — Password prompt: Translate ute → US English

🔒 Q8

Problem. Find \(F''(x)\) for \(F(x)=x^2\cos x\).

Click to Unlock Solution — Password prompt: Translate Macca’s → US English

🔒 Q9

Problem. Implicitly differentiate \(x^2+y^2=9\) and solve for \(\dfrac{dy}{dx}\).

Click to Unlock Solution — Password prompt: Translate sanga → US English

🔒 Q10

Problem. Implicitly differentiate \(e^y+xy=5\) and solve for \(\dfrac{dy}{dx}\).

Click to Unlock Solution — Password prompt: Translate brolly → US English

Frankenstein-Style Derivative Practice — Five Problems (Nested Step-by-Step Solutions)

Instructions. For each function, find \(f'(x)\) without simplifying.
Assume \(y=y(x)\) (implicit differentiation where needed).
Click to reveal Step 1, then Step 2, etc.

SOLUTIONS: may have errors - please reach out if something doesn’t seem right

Challenge Q1

\[ f_1(x)=\frac{(x^2+y)\,e^{xy}}{\,1+\sin x\,} \]

Step 1 \[ N(x)=(x^2+y)\,e^{xy},\qquad D(x)=1+\sin x. \] \[ f_1'(x)=\frac{D(x)\,N'(x)-N(x)\,D'(x)}{[D(x)]^2}. \]
Step 2 \[ N(x)=A(x)\cdot B(x),\quad A(x)=x^2+y,\; A'(x)=2x+y',\; B(x)=e^{xy},\; B'(x)=e^{xy}(y+x\,y'). \] \[ N'(x)=(2x+y')\,e^{xy}+(x^2+y)\,e^{xy}(y+x\,y'). \]
Step 3 \[ D(x)=1+\sin x\quad\Rightarrow\quad D'(x)=\cos x. \]
Step 4 \[ \boxed{f_1'(x)= \frac{(1+\sin x)\Big[(2x+y')e^{xy}+(x^2+y)e^{xy}(y+x\,y')\Big] -(x^2+y)e^{xy}\cos x}{(1+\sin x)^2}}. \]

Challenge Q2

\[ f_2(x)=\ln(x+y^2)\,\sqrt{\,1+x^2\,} \]

Step 1 \[ F(x)=\ln(x+y^2),\quad G(x)=\sqrt{\,1+x^2\,}. \] \[ f_2'(x)=F'(x)G(x)+F(x)G'(x). \]
Step 2 \[ F'(x)=\frac{1}{x+y^2}(1+2y\,y'). \]
Step 3 \[ G'(x)=\frac{x}{\sqrt{\,1+x^2\,}}. \]
Step 4 \[ \boxed{f_2'(x)= \left[\frac{1+2y\,y'}{x+y^2}\right]\sqrt{\,1+x^2\,} +\ln(x+y^2)\left[\frac{x}{\sqrt{\,1+x^2\,}}\right]}. \]

Challenge Q3

\[ f_3(x)=\frac{(x+y)\sin(x^2)}{\sqrt{\,1+y\,}} \]

Step 1 \[ N(x)=(x+y)\sin(x^2),\quad D(x)=\sqrt{\,1+y\,}. \] \[ f_3'(x)=\frac{D(x)N'(x)-N(x)D'(x)}{[D(x)]^2}. \]
Step 2 \[ C(x)=x+y,\; C'(x)=1+y',\; S(x)=\sin(x^2),\; S'(x)=2x\cos(x^2). \] \[ N'(x)=(1+y')\sin(x^2)+(x+y)2x\cos(x^2). \]
Step 3 \[ D'(x)=\frac{y'}{2\sqrt{\,1+y\,}}. \]
Step 4 \[ \boxed{f_3'(x)= \frac{\sqrt{\,1+y\,}\Big[(1+y')\sin(x^2)+(x+y)2x\cos(x^2)\Big] -(x+y)\sin(x^2)\dfrac{y'}{2\sqrt{\,1+y\,}}}{(\sqrt{\,1+y\,})^2}}. \]

Challenge Q4

\[ f_4(x)=e^{\sin(xy)}\cdot\frac{x^2+y}{\,1+xy\,} \]

Step 1 \[ P(x)=e^{\sin(xy)},\quad Q(x)=\frac{x^2+y}{1+xy}. \] \[ f_4'(x)=P'(x)Q(x)+P(x)Q'(x). \]
Step 2 \[ P'(x)=e^{\sin(xy)}\cos(xy)(y+x\,y'). \]
Step 3 \[ U(x)=x^2+y,\; U'(x)=2x+y',\; V(x)=1+xy,\; V'(x)=y+x\,y'. \] \[ Q'(x)=\frac{(2x+y')(1+xy)-(x^2+y)(y+x\,y')}{(1+xy)^2}. \]
Step 4 \[ \boxed{f_4'(x)= e^{\sin(xy)}\cos(xy)(y+x\,y')\frac{x^2+y}{1+xy} +e^{\sin(xy)}\frac{(2x+y')(1+xy)-(x^2+y)(y+x\,y')}{(1+xy)^2}}. \]

Challenge Q5

\[ f_5(x)=\sqrt{\,1+x^2y\,}\cdot\frac{\ln(x+y)}{\cos x} \]

Step 1 \[ A(x)=\sqrt{\,1+x^2y\,},\quad B(x)=\frac{\ln(x+y)}{\cos x}. \] \[ f_5'(x)=A'(x)B(x)+A(x)B'(x). \]
Step 2 \[ A'(x)=\frac{2x\,y+x^2\,y'}{2\sqrt{\,1+x^2y\,}}. \]
Step 3 \[ R(x)=\ln(x+y),\; R'(x)=\frac{1+y'}{x+y},\; S(x)=\cos x,\; S'(x)=-\sin x. \] \[ B'(x)=\frac{\frac{1+y'}{x+y}\cos x+\ln(x+y)\sin x}{\cos^2 x}. \]
Step 4 \[ \boxed{f_5'(x)= \frac{2x\,y+x^2\,y'}{2\sqrt{\,1+x^2y\,}}\cdot\frac{\ln(x+y)}{\cos x} +\sqrt{\,1+x^2y\,}\left[\frac{\frac{1+y'}{x+y}\cos x+\ln(x+y)\sin x}{\cos^2 x}\right]}. \]

Test 2 — Practice Test II

Instructions. Each problem has a click-to-reveal, step-by-step solution.
Assume \(y=y(x)\) when a problem uses implicit differentiation.
You do not need to simplify final derivatives unless stated.

SOLUTIONS: may have errors - please reach out if something doesn’t seem right

Q1

\[ F(x)=\frac{3}{5}x^{5}-4x^{1/2}+7. \]

Step-by-Step Solution

\[ \frac{d}{dx}\!\left(\tfrac{3}{5}x^{5}\right)=3x^{4},\qquad \frac{d}{dx}\!\left(-4x^{1/2}\right)=-2x^{-1/2},\qquad \frac{d}{dx}(7)=0. \] \[ \boxed{F'(x)=3x^{4}-2x^{-1/2}.} \]

Q2

\[ F(x)=(x^{2}+1)\ln x \quad (x>0) \]

Step-by-Step Solution

Product rule: \(f'(x)=2x,\ g'(x)=1/x.\) \[ F'(x)=f'g+fg' = 2x\ln x+(x^{2}+1)\frac1x = 2x\ln x+x+\frac1x. \] \[ \boxed{F'(x)=2x\ln x+x+\frac1x.} \]

Q3

\[ F(x)=\frac{e^{2x}}{1+x}. \]

Step-by-Step Solution

\(f'(x)=2e^{2x},\ g'(x)=1.\) \[ F'(x)=\frac{f'g-fg'}{g^{2}} =\frac{2e^{2x}(1+x)-e^{2x}}{(1+x)^{2}} =\frac{e^{2x}(1+2x)}{(1+x)^2}. \] \[ \boxed{F'(x)=\frac{e^{2x}(1+2x)}{(1+x)^2}.} \]

Q4

\[ F(x)=\sin\!\big(\ln(1+x^{2})\big). \]

Step-by-Step Solution

Chain rule:
Outer \(\sin(\cdot)\to\cos(\cdot)\cdot(\cdot)'\).
Inner \(h(x)=\ln(1+x^{2})\Rightarrow h'(x)=\frac{2x}{1+x^{2}}\). \[ F'(x)=\cos\!\big(\ln(1+x^{2})\big)\cdot\frac{2x}{1+x^{2}}. \] \[ \boxed{F'(x)=\frac{2x}{1+x^{2}}\cos\!\big(\ln(1+x^{2})\big).} \]

Q5

\[ F(x)=x\,e^{\cos x}. \]

Step-by-Step Solution

Product rule with \(f'(x)=1,\ g'(x)=e^{\cos x}(-\sin x)\): \[ F'(x)=f'g+fg'=e^{\cos x}+x\,e^{\cos x}(-\sin x)=e^{\cos x}(1-x\sin x). \] \[ \boxed{F'(x)=e^{\cos x}(1-x\sin x).} \]

Q6

\[ F(x)=\ln(1+x^{3}). \]

Step-by-Step Solution

Chain rule — outer \(\ln(\cdot)\), inner \(1+x^{3}\): \[ F'(x)=\frac{1}{1+x^{3}}\cdot(3x^{2})=\frac{3x^{2}}{1+x^{3}}. \] \[ \boxed{F'(x)=\frac{3x^{2}}{1+x^{3}}.} \]

Q7

\[ x\,e^{y}+y^{2}=1. \]

Step-by-Step Solution

Differentiate: \[ e^{y}+x\,e^{y}y'+2y\,y'=0. \] \[ y'(x\,e^{y}+2y)=-e^{y}. \] \[ \boxed{y'=-\frac{e^{y}}{x\,e^{y}+2y}.} \]

Q8

\[ \sin(xy)=x. \]

Step-by-Step Solution

Differentiate: \[ \cos(xy)(y+x\,y')=1. \] \[ x\cos(xy)y'=1-\cos(xy)y. \] \[ \boxed{y'=\frac{1-\cos(xy)y}{x\cos(xy)}.} \]

Q9

\[ F(x)=x\,e^{x}. \]

Step-by-Step Solution

First derivative: \[ F'(x)=e^{x}(1+x). \] Second derivative: \[ F''(x)=e^{x}(1+x)+e^{x}=e^{x}(x+2). \] \[ \boxed{F''(x)=e^{x}(x+2).} \]

Q10

\[ F(x)=\frac{\ln(1+x e^{x})}{\sqrt{1-x^{2}}},\quad |x|<1. \]

Step-by-Step Solution

\(f(x)=\ln(1+x e^{x})\Rightarrow f'(x)=\dfrac{e^{x}(1+x)}{1+x e^{x}}\).
\(g(x)=\sqrt{1-x^{2}}\Rightarrow g'(x)=-\dfrac{x}{\sqrt{1-x^{2}}}.\)

\[ F'(x)=\frac{f'g-fg'}{g^{2}} =\frac{\frac{e^{x}(1+x)}{1+x e^{x}}\sqrt{1-x^{2}}+\frac{x\ln(1+x e^{x})}{\sqrt{1-x^{2}}}}{1-x^{2}}. \] \[ \boxed{F'(x)= \frac{\frac{e^{x}(1+x)}{1+x e^{x}}\sqrt{1-x^{2}}+\frac{x\ln(1+x e^{x})}{\sqrt{1-x^{2}}}} {1-x^{2}}.} \]