All Calculus 1 Resources
Example Questions
Example Question #1 : Other Differential Functions
What is the first derivative of sin(x2)(3x2 + 5x)?
6x * sin(x2)+ 5sin(x2) - 6x3 * cos(x2) + 10x2 * cos(x2)
None of the other answers
6x * cos(x) + 10 * cos(x)
6 * sin(x2) + 5 * sin(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)
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)
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))?
2x * tan(x2) + x2 * 2x * sec2(x2)
(2/x) + 2x * sec(x2) * csc(x2)
2x * sec2(x2)
1/(2x * tan(x2) + x2 * 2x * sec2(x2))
1/(x2 * tan(x2))
(2/x) + 2x * sec(x2) * csc(x2)
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 #3 : How To Find Differential Functions
What is the first derivative of s(x) = 2sin(sin(x))?
ln(2) * 2cos(sin(x)) * cos(x)
ln(2) * 2sin(sin(x)) * cos(sin(x)) * cos(x)
2cos(x)sin(x)
2cos(sin(x)) * cos(x)
None of the other answers
ln(2) * 2sin(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) = 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 #2 : Other Differential Functions
What is the first derivative of f(x) = sin2(x) – sin(x)cos(x)?
2sin(x)cos(x) – 1
sin2(x) – 2sin(x)cos(x) – cos2(x)
2sin(x)cos(x) + cos2(x) + sin2(x)
2sin(x)cos(x) – cos2(x) + sin2(x)
2sin(x)cos(x) + 1
2sin(x)cos(x) – cos2(x) + sin2(x)
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 : Other Differential Functions
What is the first derivative of f(x) = sin(x)ln(cos(x))?
cos(x)ln(cos(x)) – tan(x)csc(x)
cos(x)ln(cos(x)) + tan(x)sin(x)
cos(x)ln(cos(x)) – tan(x)sin(x)
None of the other answers
cos(x)ln(sin(x)) – sin2(x)
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)) = –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 #5 : How To Find Differential Functions
What is the first derivative of f(x) = sin(cos(tan(sin(x))))
–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))
None of the other answers
sin(sin(tan(sin(x)))) * tan(sin(x))cos(x) * sin(sec2(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)))
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 #191 : Differential Functions
What is the first derivative of f(x) = sec(x2 + 4x)?
sec(x2 + 4x)tan(x2 + 4x)
2x * tan2(x2 + 4x) + 4tan2(x2 + 4x)
2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)
2x * tan(x2 + 4x) + 4tan(x2 + 4x)
sec(2x + 4)tan(2x + 4)
2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)
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 #2 : How To Find Differential Functions
What is the first derivative of f(x) = cos4(x2)
4x * cos3(x2)
–8x * sin(x2)cos3(x2)
4cos3(x2)
4x * sin(x2)cos3(x2)
4cos3(x2)sin(x2)
–8x * sin(x2)cos3(x2)
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 #1 : How To Find Differential Functions
What is the first derivative of f(x) = (100/x2) + (50/x) – 200x2?
(50 + 50x – 400x2)/x
None of the other answers
(14 + 4x + 3x4)/x3
–50(4 + x + 8x4)/x3
200x + 50 – 400x
–50(4 + x + 8x4)/x3
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
Certified Tutor
Certified Tutor