### All Calculus 1 Resources

## Example Questions

### Example Question #1217 : Calculus

What is the first derivative of sin(x^{2})(3x^{2} + 5x)?

**Possible Answers:**

6 * sin(x^{2}) + 5 * sin(x^{2})

6x * cos(x) + 10 * cos(x)

6x * sin(x^{2})+ 5sin(x^{2}) - 6x^{3} * cos(x^{2}) + 10x^{2} * cos(x^{2})

None of the other answers

6x^3 * cos(x^{2}) + 10x^{2} * cos(x^{2}) + 6x * sin(x^{2})+ 5sin(x^{2})

**Correct answer:**

6x^3 * cos(x^{2}) + 10x^{2} * cos(x^{2}) + 6x * sin(x^{2})+ 5sin(x^{2})

To find the derivative, we must use both the product and the chain rule. Let us consider each part:

f(x) = sin(x^{2})

g(x) = 3x^{2} + 5x

For f(x), we will need to apply the chain rule. Consider f(x) as being f(h(x)) where h(x) = x^{2;} therefore f'(x) = f'(h(x)) * h'(x) or cos(x^{2}) * 2x = 2x * cos(x^{2}).

Gathering together all of our data, we have:

f(x) = sin(x^{2})

f'(x) = 2x * cos(x^{2})

g(x) = 3x^{2} + 5x

g'(x) = 6x + 5

The product rule states that if a(x) = f(x) * g(x), then a'(x) = f'(x) * g(x) + f(x) * g'(x)

For our data, then, the derivative of a(x) = sin(x^{2})(3x^{2} + 5x) is:

a'(x) = 2x * cos(x^{2}) * (3x^{2} + 5x) + sin(x^{2}) * (6x + 5)

Distributing, this gives us:

a'(x) = 6x^{3}cos(x^{2}) + 10x^{2}cos(x^{2}) + 6x * sin(x^{2})+ 5sin(x^{2})

### Example Question #1218 : Calculus

What is the first derivative of f(x) = sin(2x^{2} + 5x) – sin(cos(x))?

**Possible Answers:**

sin(4x + 5 + sin(x))

cos(2x^{2} + 5x) + cos(cos(x))

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

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

sin(4x + 5) + sin(–sin(x))

**Correct answer:**

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

For each of the parts of f(x), you will have to apply the chain rule:

For sin(2x^{2} + 5x): cos(2x^{2} + 5x) * (4x + 5)

For (sin(cos(x)): cos(cos(x))*(-sin(x))

Therefore, for our original:

f'(x) = cos(2x^{2} + 5x) * (4x + 5) – cos(cos(x)) * (–sin(x))

Simplifying:

f'(x) = 4x*cos(2x^{2} + 5x) + 5cos(2x^{2} + 5x) + cos(cos(x)) * sin(x)

### Example Question #181 : Differential Functions

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

**Possible Answers:**

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

1/(2x * tan(x^{2}) + x^{2} * 2x * sec^{2}(x^{2}))

1/(x^{2} * tan(x^{2}))

(2/x) + 2x * sec(x^{2}) * csc(x^{2})

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

**Correct answer:**

(2/x) + 2x * sec(x^{2}) * csc(x^{2})

f(x) = ln(x^{2 *} tan(x^{2}))?

The chain rule is necessary. For the natural logarithm "portion", we know the derivative will be:1/(x^{2} * tan(x^{2}))

However, we must add the derivative of the argument using the product rule: 2x*tan(x^{2}) + x^{2} * 2x * sec^{2}(x^{2})

The whole derivative: f'(x) = (2x * tan(x^{2}) + x^{2} * 2x * sec^{2}(x^{2}))/(x^{2} * tan(x^{2}))

Splitting apart the fraction, we have the sum of:

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

and

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

The first simplifies to 2/x

The second is a bit trickier. First let us simplify the non-trigonometric components:

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

Now, since sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), we can rewrite:

2x * sec^{2}(x^{2})/tan(x^{2}) = 2x/(cos^{2}(x^{2}) * sin(x^{2})/cos(x^{2}))

This reduces to: 2x/(cos(x^{2}) * sin(x^{2})), which is the same as 2x * sec(x^{2}) * csc(x^{2})

Therefore, we can write our solution as f'(x) = (2/x) + 2x * sec(x^{2}) * csc(x^{2})

### Example Question #1220 : Calculus

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

**Possible Answers:**

ln(2) * 2^{cos(sin(x)) * cos(x)}

2^{cos(sin(x)) * cos(x)}

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

ln(2) * 2^{sin(sin(x))} * cos(sin(x)) * cos(x)

None of the other answers

**Correct answer:**

ln(2) * 2^{sin(sin(x))} * cos(sin(x)) * cos(x)

We must carefully parse this function in order to apply the chain rule correctly. Let us make the following substitutions:

f(x) = 2^{x}

g(x) = sin(sin(x))

s(x) = f(g(x)) = 2^{sin(}^{sin(x))}

According to the chain rule, we know that s'(x) = f'(g(x)) * g'(x)

f'(g(x)) = ln(2) * 2^{sin(sin(x))}

g'(x) = cos(sin(x)) * cos(x)

Therefore, s'(x) = ln(2) * 2^{sin(sin(x))} * cos(sin(x)) * cos(x)

### Example Question #1 : How To Find Differential Functions

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

**Possible Answers:**

2sin(x)cos(x) – cos^{2}(x) + sin^{2}(x)

2sin(x)cos(x) + 1

2sin(x)cos(x) – 1

sin^{2}(x) – 2sin(x)cos(x) – cos^{2}(x)

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

**Correct answer:**

2sin(x)cos(x) – cos^{2}(x) + sin^{2}(x)

Consider each of the elements in isolation:

sin^{2}(x) can be differentiated using the chain rule.

Step 1: 2(sin(x))

Step 2: take the derivative of sin(x): cos(x)

Therefore, d/dx (sin^{2}(x)) = 2sin(x)cos(x)

sin(x)cos(x) can be differentiated by the product rule:

cos(x)cos(x) + sin(x) * –sin(x) = cos^{2}(x) – sin^{2}(x)

Combine these:

2sin(x)cos(x) – (cos^{2}(x) – sin^{2}(x)) = 2sin(x)cos(x) – cos^{2}(x) + sin^{2}(x)

### Example Question #2 : How To Find Differential Functions

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

**Possible Answers:**

cos(x)ln(cos(x)) + tan(x)sin(x)

None of the other answers

cos(x)ln(sin(x)) – sin^{2}(x)

cos(x)ln(cos(x)) – tan(x)csc(x)

cos(x)ln(cos(x)) – tan(x)sin(x)

**Correct answer:**

cos(x)ln(cos(x)) – tan(x)sin(x)

This is a mixture of the product rule and the chain rule:

The first term of the product rule is: cos(x)ln(cos(x))

The second term will have sin(x) but will include the differentiation of the ln(cos(x)), which will require the chain rule:

sin(x) * (1/cos(x)) * (–sin(x)) = –sin^{2}(x)/cos(x)

Combining both we get:

cos(x)ln(cos(x)) – sin^{2}(x)/cos(x)

Now, note that none of the answers are the same as this ;however, we can make an alteration:

sin(x)/cos(x) is the same as tan(x)

Therefore, the answer is: cos(x)ln(cos(x)) – tan(x)sin(x)

### Example Question #3 : How To Find Differential Functions

What is the first derivative of f(x) = sin(cos(tan(sin(x))))

**Possible Answers:**

–cos(cos(tan(sin(x)))) * sec^{2}(sin(x))cos(x) * sin(tan(sin(x)))

cos(cos(tan(sin(x))))

cos(cos(tan(sin(x))))) * cos(tan(sin(x)))

sin(sin(tan(sin(x)))) * tan(sin(x))cos(x) * sin(sec^{2}(sin(x)))

None of the other answers

**Correct answer:**

–cos(cos(tan(sin(x)))) * sec^{2}(sin(x))cos(x) * sin(tan(sin(x)))

Okay, don't be overwhelmed. Take this chain rule one step at a time:

Step 1: Do the sine...

cos(cos(tan(sin(x))))

Step 2: Do the cosine . . .

–sin(tan(sin(x)))

Step 3: do the tangent . . . this is the simple chain rule, so diffentiate the argument as well

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

Step 4: Multiply them together:

–cos(cos(tan(sin(x)))) * sec^{2}(sin(x))cos(x) * sin(tan(sin(x)))

### Example Question #4 : How To Find Differential Functions

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

**Possible Answers:**

2x * tan(x^{2} + 4x) + 4tan(x^{2} + 4x)

sec(2x + 4)tan(2x + 4)

sec(x^{2} + 4x)tan(x^{2} + 4x)

2x * sec(x^{2} + 4x)tan(x^{2} + 4x) + 4sec(x^{2} + 4x)tan(x^{2} + 4x)

2x * tan^{2}(x^{2} + 4x) + 4tan^{2}(x^{2} + 4x)

**Correct answer:**

2x * sec(x^{2} + 4x)tan(x^{2} + 4x) + 4sec(x^{2} + 4x)tan(x^{2} + 4x)

This is a simple chain rule. The derivative of the secant is secant * tangent; therefore:

f'(x) = sec(x^{2} + 4x) * tan(x^{2} + 4x) * (2x + 4)

Distribute everything to get your answer: 2x * sec(x^{2} + 4x)tan(x^{2} + 4x) + 4sec(x^{2} + 4x)tan(x^{2} + 4x)

### Example Question #1 : How To Find Differential Functions

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

**Possible Answers:**

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

–8x * sin(x^{2})cos^{3}(x^{2})

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

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

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

**Correct answer:**

–8x * sin(x^{2})cos^{3}(x^{2})

Consider this as a chain rule case. Do each step:

Step 1: cos^{4}

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

Step 2: cos(x^{2}); this can be treated like a normal case of the chain rule

–sin(x^{2}) * 2x

Combining these, we get

–8x * sin(x^{2})cos^{3}(x^{2})

### Example Question #6 : How To Find Differential Functions

What is the first derivative of f(x) = (100/x^{2}) + (50/x) – 200x^{2}?

**Possible Answers:**

None of the other answers

(14 + 4x + 3x^{4})/x^{3}

–50(4 + x + 8x^{4})/x^{3}

200x + 50 – 400x

(50 + 50x – 400x^{2})/x

**Correct answer:**

–50(4 + x + 8x^{4})/x^{3}

f(x) = (100/x^{2}) + (50/x) – 200x^{2}

First, rewrite the equation: 100x^{–2} + 50x^{–1} – 200x^{2}

At this point, it is relatively easy to differentiate:

f'(x) = –2 * 100x^{–3} – 50x^{–2} – 400x = (–200/x^{3}) – (50/x^{2}) – 400x

Simplify by making x^{3} the common denominator:

(–200 – 50x – 400x^{4})/x^{3}

Factor out the common –50 in the numerator to make things look nicer:

–50(4 + x + 8x^{4})/x^{3}