### All Calculus 1 Resources

## Example Questions

### Example Question #1221 : Calculus

What is the first derivative of f(x) = 2x * ln(sin(x))?

**Possible Answers:**

2(ln(sin(x)) + x * tan(x))

2(ln(sin(x)) + x * cot(x))

2tan(x)

None of the other answers

2ln(sin(x))/cos(x)

**Correct answer:**

2(ln(sin(x)) + x * cot(x))

f(x) = 2x * ln(sin(x))

Here, we have a product rule with a chain rule.

The derivative of ln(sin(x)) = (1/sin(x)) * cos(x). Note that this is cot(x).

Therefore, our full derivative is:

f'(x) = 2ln(sin(x)) + 2x * cot(x) = 2(ln(sin(x)) + x * cot(x))

### Example Question #1222 : Calculus

What is the first derivative of f(x) = sin^{4}(x) – cos^{3}(x)?

**Possible Answers:**

sin(x) + 1

4sin^{3}(x)cos(x) + 3cos^{2}(x)sin(x)

4sin^{3}(x) + 3cos^{2}(x)

sin^{4}(x)cos(x) + cos^{3}(x)sin(x)^{}

4sin^{3}(x)cos(x) – 3cos^{2}(x)sin(x)

**Correct answer:**

4sin^{3}(x)cos(x) + 3cos^{2}(x)sin(x)

f(x) = sin^{4}(x) – cos^{3}(x)?

Treat these as chain rules:

The derivative of sin^{4}(x) is therefore: 4sin^{3}(x)cos(x)

The derivative of cos^{3}(x) is therefore: 3cos^{2}(x) * –sin(x) or –3cos^{2}(x)sin(x)

This gives us 4sin^{3}(x)cos(x) + 3cos^{2}(x)sin(x), which really cannot be simplified any further.

### Example Question #1223 : Calculus

What is the the first derivative of f(x) = 2^{tan(x)}?^{}

**Possible Answers:**

ln(2) * 2^{tan(x)}

cot(x) * ln(2) * 2^{tan(x)}

ln(2) * 2^{sec(x)tan}^{(x)}

sec(x) * tan(x) * ln(2) * 2^{tan(x)}

sec^{2}(x) * ln(2) * 2^{tan(x)}

**Correct answer:**

sec^{2}(x) * ln(2) * 2^{tan(x)}

This is a case of the chain rule.

Step 1: Deal with the exponential function

ln(2) * 2^{tan(x)}

Step 2: Deal with the exponent

sec^{2}(x)

Multiply them together to get your answer:

sec^{2}(x) * ln(2) * 2^{tan(x)}

### Example Question #1224 : Calculus

What is the first derivative of f(x) = sin^{4}(x) – cos^{4}(x)?

**Possible Answers:**

4

1

2sin(2x)

4sin^{3}(x) + 4cos^{3}(x)

4sin(4x)

**Correct answer:**

2sin(2x)

Applying the chain rule to each element, we get:

f'(x) = 4sin^{3}(x)cos(x) + 4cos^{3}(x)sin(x)

If we factor out the common factors, we get:

f'(x) = 4sin(x)cos(x)(sin^{2}(x) + cos^{2}(x)) = 4sin(x)cos(x)(1) = 4sin(x)cos(x)

Also, since we know that 2sin(x)cos(x) = sin(2x), we know that 4sin(x)cos(x) = 2sin(2x)

### Example Question #15 : Other Differential Functions

What is the first derivative of f(x) = x^{2}sin(4x^{3})?

**Possible Answers:**

2x * sin(4x^{3}) + 12x^{5}cos(4x^{3})

sin(4x^{3}) + 12x^{2}cos(4x^{3})

2x(sin(4x^{3}) + 6x^{3}cos(4x^{3}))

2x * cos(4x^{3})

24x^{4} * cos(4x^{3})

**Correct answer:**

2x(sin(4x^{3}) + 6x^{3}cos(4x^{3}))

This is a product rule combined with a chain rule. Let's do the chain rule for sin(4x^{3}) first:

cos(4x^{3}) * 12x^{2} = 12x^{2}cos(4x^{3})

With this in mind, let's solve our whole problem:

2x * sin(4x^{3}) + x^{2} * 12x^{2}cos(4x^{3}) = 2x * sin(4x^{3}) + 12x^{4}cos(4x^{3}) = 2x(sin(4x^{3}) + 6x^{3}cos(4x^{3}))

### Example Question #16 : Other Differential Functions

What is the first derivative of f(x) = x^{4} – x * sin(x^{–5})?

**Possible Answers:**

4x^{3} + cos(x^{–5})

4x^{3} – cos(x^{–5})

None of the other answers

4x^{3} – sin(x^{–5}) + (5 * cos(x^{–5})/x^{5})

4x^{3} + sin(x^{–5}) – (5 * cos(x^{–5})/x^{5})

**Correct answer:**

4x^{3} – sin(x^{–5}) + (5 * cos(x^{–5})/x^{5})

The first element is merely differentiated as 4x^{3}

The second element is a relatively simple product rule:

sin(x^{–5}) + x * cos(x^{–5}) * –5 * x^{–6} = sin(x^{–5}) + cos(x^{–5}) * –5 * x^{–5} = sin(x^{–5}) – (5 * cos(x^{–5})/x^{5})

Put everything back together:

4x^{3} – ( sin(x^{–5}) – (5 * cos(x^{–5})/x^{5}) ) = 4x^{3} – sin(x^{–5}) + (5 * cos(x^{–5})/x^{5})

### Example Question #11 : How To Find Differential Functions

What is the first derivative of f(x) = 5x * ln(2x)?

**Possible Answers:**

5(ln(2) + 1)

5/2x

5ln(2)/2x

5(ln(2) + (1/2))

5/x

**Correct answer:**

5(ln(2) + 1)

This is just a normal product rule problem:

5 * ln(2x) + 5x * 2 * (1/2x)

Simplify: 5ln(2) + 5 = 5(ln(2) + 1)

### Example Question #15 : Other Differential Functions

What is the slope of the tangent line at x = 2 for f(x) = 6/x^{2}?

**Possible Answers:**

–1.5

1.5

3

–12

–3

**Correct answer:**

–1.5

First rewrite your function to make this easier:

f(x) = 6/x^{2 }= 6x^{–2}

Now, we must find the first derivative:

f'(x) = –2 * 6 * x^{–3} = –12/x^{3}^{}

The slope of the tangent line of f(x) at x = 2 is: f'(2) = –12/2^{3} = –12/8 = –3/2 = –1.5

### Example Question #16 : How To Find Differential Functions

What is the first derivative of f(x) = 2ln(cos(x)sin(x))?

**Possible Answers:**

2cot(x)cos^{2}(x)

cot(x)cos^{2}(x)

None of the other answers

2cos^{2}(x)sec(x)csc(x)

2cos(2x)sec(x)csc(x)

**Correct answer:**

2cos(2x)sec(x)csc(x)

This requires both the use of the chain rule and the product rule. Start with the natural logarithm: 2/(cos(x)sin(x))

Now, multiply by d/dx cos(x)sin(x), which is: –sin(x)sin(x) + cos(x)cos(x) = cos^{2}(x) – sin^{2}(x)

Therefore, f'(x) = (cos^{2}(x) – sin^{2}(x)) * 2/(cos(x)sin(x)) = 2(cos^{2}(x) – sin^{2}(x))sec(x)csc(x)

From our trigonometric identities, we know cos^{2}(x) – sin^{2}(x) = cos(2x)

Therefore, we can finilize our simplification to f'(x) = 2cos(2x)sec(x)csc(x)

### Example Question #11 : Other Differential Functions

Take the derivative of

**Possible Answers:**

Derivative does not exist

**Correct answer:**

We need to use logarithm differentiation to do this problem. Take the natural log of both sides.

Apply the power rule of natural log

Perform implicit differentiation to both sides

Solve for