# Calculus 1 : How to find differential functions

## Example Questions

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### Example Question #1211 : Calculus

What is the first derivative of sin(x2)(3x2 + 5x)?

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

6 * sin(x2) + 5 * sin(x2)

6x * sin(x2)+ 5sin(x2) - 6x3 * cos(x2) + 10x2 * cos(x2)

6x^3 * cos(x2) + 10x2 * cos(x2) + 6x * sin(x2)+ 5sin(x2)

6x^3 * cos(x2) + 10x2 * cos(x2) + 6x * sin(x2)+ 5sin(x2)

Explanation:

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

f(x) = sin(x2)

g(x) = 3x2 + 5x

For f(x), we will need to apply the chain rule.  Consider f(x) as being f(h(x)) where h(x) = x2; therefore f'(x) = f'(h(x)) * h'(x) or cos(x2) * 2x = 2x * cos(x2).

Gathering together all of our data, we have:

f(x) = sin(x2)

f'(x) = 2x * cos(x2)

g(x) = 3x2 + 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(x2)(3x2 + 5x) is:

a'(x) = 2x * cos(x2) * (3x2 + 5x) + sin(x2) * (6x + 5)

Distributing, this gives us:

a'(x) = 6x3cos(x2) + 10x2cos(x2) + 6x * sin(x2)+ 5sin(x2)

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

What is the first derivative of f(x) = sin(2x2 + 5x) – sin(cos(x))?

4x*cos(2x2 + 5x) + 5cos(2x2 + 5x) + cos(cos(x)) * sin(x)

4*cos(2x2 + 5x) + 5cos(2x2 + 5x) - cos(cos(x)) * sin(x)

cos(2x2 + 5x) + cos(cos(x))

sin(4x + 5) + sin(–sin(x))

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

4x*cos(2x2 + 5x) + 5cos(2x2 + 5x) + cos(cos(x)) * sin(x)

Explanation:

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

For sin(2x2 + 5x): cos(2x2 + 5x) * (4x + 5)

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

Therefore, for our original:

f'(x) = cos(2x2 + 5x) * (4x + 5) – cos(cos(x)) * (–sin(x))

Simplifying:

f'(x) = 4x*cos(2x2 + 5x) + 5cos(2x2 + 5x) + cos(cos(x)) * sin(x)

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

What is the first derivative of f(x) = ln(x2 tan(x2))?

(2/x) + 2x * sec(x2) * csc(x2)

1/(2x * tan(x2) + x2 * 2x * sec2(x2))

2x * tan(x2) + x2 * 2x * sec2(x2)

2x * sec2(x2)

1/(x2 * tan(x2))

(2/x) + 2x * sec(x2) * csc(x2)

Explanation:

f(x) = ln(x2 * tan(x2))?

The chain rule is necessary.  For the natural logarithm "portion", we know the derivative will be:1/(x2 * tan(x2))

However, we must add the derivative of the argument using the product rule: 2x*tan(x2) + x2 * 2x * sec2(x2)

The whole derivative: f'(x) = (2x * tan(x2) + x2 * 2x * sec2(x2))/(x2 * tan(x2))

Splitting apart the fraction, we have the sum of:

(2x * tan(x2))/(x2 * tan(x2))

and

x2 * 2x * sec2(x2)/(x2 * tan(x2))

The first simplifies to 2/x

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

2x * sec2(x2)/tan(x2)

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

2x * sec2(x2)/tan(x2) = 2x/(cos2(x2) * sin(x2)/cos(x2))

This reduces to: 2x/(cos(x2) * sin(x2)), which is the same as 2x * sec(x2) * csc(x2)

Therefore, we can write our solution as f'(x) = (2/x) + 2x * sec(x2) * csc(x2)

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

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

2cos(sin(x)) * cos(x)

ln(2) * 2cos(sin(x)) * cos(x)

2cos(x)sin(x)

ln(2) * 2sin(sin(x))  * cos(sin(x)) * cos(x)

ln(2) * 2sin(sin(x))  * cos(sin(x)) * cos(x)

Explanation:

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

f(x) = 2x

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

s(x) = f(g(x)) = 2sin(sin(x))

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

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

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

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

### Example Question #5 : How To Find Differential Functions

What is the first derivative of f(x) = sin2(x) – sin(x)cos(x)?

2sin(x)cos(x) – 1

2sin(x)cos(x) + cos2(x) + sin2(x)

2sin(x)cos(x) – cos2(x) + sin2(x)

2sin(x)cos(x) + 1

sin2(x) – 2sin(x)cos(x) – cos2(x)

2sin(x)cos(x) – cos2(x) + sin2(x)

Explanation:

Consider each of the elements in isolation:

sin2(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 (sin2(x)) = 2sin(x)cos(x)

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

cos(x)cos(x) + sin(x) * –sin(x) = cos2(x) – sin2(x)

Combine these:

2sin(x)cos(x) – (cos2(x) – sin2(x)) = 2sin(x)cos(x) – cos2(x) + sin2(x)

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

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

cos(x)ln(sin(x)) – sin2(x)

cos(x)ln(cos(x)) – tan(x)sin(x)

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

cos(x)ln(cos(x)) – tan(x)csc(x)

cos(x)ln(cos(x)) – tan(x)sin(x)

Explanation:

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)) = –sin2(x)/cos(x)

Combining both we get:

cos(x)ln(cos(x)) – sin2(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 #1 : How To Find Differential Functions

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

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

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

–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))

sin(sin(tan(sin(x)))) * tan(sin(x))cos(x) * sin(sec2(sin(x)))

–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))

Explanation:

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

sec2(sin(x))cos(x)

Step 4: Multiply them together:

–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))

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

What is the first derivative of f(x) = sec(x2 + 4x)?

2x * tan(x2 + 4x) + 4tan(x2 + 4x)

2x * tan2(x2 + 4x) + 4tan2(x2 + 4x)

sec(x2 + 4x)tan(x2 + 4x)

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

2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)

2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)

Explanation:

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

f'(x) = sec(x2 + 4x) * tan(x2 + 4x) * (2x + 4)

Distribute everything to get your answer: 2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)

### Example Question #1 : Other Differential Functions

What is the first derivative of f(x) = cos4(x2)

4cos3(x2)sin(x2)

4cos3(x2)

–8x * sin(x2)cos3(x2)

4x * sin(x2)cos3(x2)

4x * cos3(x2)

–8x * sin(x2)cos3(x2)

Explanation:

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

Step 1: cos4

4cos3(x2)

Step 2: cos(x2); this can be treated like a normal case of the chain rule

–sin(x2) * 2x

Combining these, we get

–8x * sin(x2)cos3(x2)

### Example Question #10 : How To Find Differential Functions

What is the first derivative of f(x) = (100/x2) + (50/x) – 200x2?

200x + 50 – 400x

–50(4 + x + 8x4)/x3

(14 + 4x + 3x4)/x3

(50 + 50x – 400x2)/x

–50(4 + x + 8x4)/x3

Explanation:

f(x) = (100/x2) + (50/x) – 200x2

First, rewrite the equation: 100x–2 + 50x–1 – 200x2

At this point, it is relatively easy to differentiate:

f'(x) = –2 * 100x–3 – 50x–2 – 400x = (–200/x3) – (50/x2) – 400x

Simplify by making x3 the common denominator:

(–200 – 50x – 400x4)/x3

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

–50(4 + x + 8x4)/x3

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