All Calculus 1 Resources
Example Questions
Example Question #201 : Differential Functions
What is the first derivative of f(x) = sin4(x) – cos4(x)?
4sin(4x)
4sin3(x) + 4cos3(x)
1
2sin(2x)
4
2sin(2x)
Applying the chain rule to each element, we get:
f'(x) = 4sin3(x)cos(x) + 4cos3(x)sin(x)
If we factor out the common factors, we get:
f'(x) = 4sin(x)cos(x)(sin2(x) + cos2(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 #202 : Differential Functions
What is the first derivative of f(x) = x2sin(4x3)?
sin(4x3) + 12x2cos(4x3)
2x(sin(4x3) + 6x3cos(4x3))
2x * cos(4x3)
24x4 * cos(4x3)
2x * sin(4x3) + 12x5cos(4x3)
2x(sin(4x3) + 6x3cos(4x3))
This is a product rule combined with a chain rule. Let's do the chain rule for sin(4x3) first:
cos(4x3) * 12x2 = 12x2cos(4x3)
With this in mind, let's solve our whole problem:
2x * sin(4x3) + x2 * 12x2cos(4x3) = 2x * sin(4x3) + 12x4cos(4x3) = 2x(sin(4x3) + 6x3cos(4x3))
Example Question #203 : Differential Functions
What is the first derivative of f(x) = x4 – x * sin(x–5)?
4x3 – cos(x–5)
None of the other answers
4x3 + cos(x–5)
4x3 + sin(x–5) – (5 * cos(x–5)/x5)
4x3 – sin(x–5) + (5 * cos(x–5)/x5)
4x3 – sin(x–5) + (5 * cos(x–5)/x5)
The first element is merely differentiated as 4x3
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)/x5)
Put everything back together:
4x3 – ( sin(x–5) – (5 * cos(x–5)/x5) ) = 4x3 – sin(x–5) + (5 * cos(x–5)/x5)
Example Question #204 : Differential Functions
What is the first derivative of f(x) = 5x * ln(2x)?
5ln(2)/2x
5/x
5(ln(2) + (1/2))
5/2x
5(ln(2) + 1)
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 #204 : Differential Functions
What is the slope of the tangent line at x = 2 for f(x) = 6/x2?
–1.5
–3
–12
1.5
3
–1.5
First rewrite your function to make this easier:
f(x) = 6/x2 = 6x–2
Now, we must find the first derivative:
f'(x) = –2 * 6 * x–3 = –12/x3
The slope of the tangent line of f(x) at x = 2 is: f'(2) = –12/23 = –12/8 = –3/2 = –1.5
Example Question #205 : Differential Functions
What is the first derivative of f(x) = 2ln(cos(x)sin(x))?
2cot(x)cos2(x)
cot(x)cos2(x)
2cos(2x)sec(x)csc(x)
2cos2(x)sec(x)csc(x)
None of the other answers
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) = cos2(x) – sin2(x)
Therefore, f'(x) = (cos2(x) – sin2(x)) * 2/(cos(x)sin(x)) = 2(cos2(x) – sin2(x))sec(x)csc(x)
From our trigonometric identities, we know cos2(x) – sin2(x) = cos(2x)
Therefore, we can finilize our simplification to f'(x) = 2cos(2x)sec(x)csc(x)
Example Question #206 : Differential Functions
Take the derivative of
Derivative does not exist
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
Example Question #21 : Other Differential Functions
Find
Example Question #1237 : Calculus
Differentiate:
with respect to .
Apply the chain rule: differentiate the "outside" function first. Let .
Differentiate the "inside" function next.
Multiply these two functions to find the derivative of the original function.
Example Question #1238 : Calculus
Evaluate
To integrate the function, integrate each term of the function. e.g., integrate by increasing the exponent by 1 integer and dividing the term by this new integer: .
Do this for the rest to get .
But remember that every integration requires an arbitrary constant, . Thus, the integral of the function is