Calculus 1 : How to find differential functions
Study concepts, example questions & explanations for Calculus 1
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Example Questions
Example Question #11 : Other Differential Functions
What is the first derivative of f(x) = 2x * ln(sin(x))?
2ln(sin(x))/cos(x)
2tan(x)
None of the other answers
2(ln(sin(x)) + x * tan(x))
2(ln(sin(x)) + x * cot(x))
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 #12 : How To Find Differential Functions
What is the first derivative of f(x) = sin4(x) – cos3(x)?
sin(x) + 1
4sin3(x)cos(x) – 3cos2(x)sin(x)
4sin3(x)cos(x) + 3cos2(x)sin(x)
sin4(x)cos(x) + cos3(x)sin(x)
4sin3(x) + 3cos2(x)
4sin3(x)cos(x) + 3cos2(x)sin(x)
f(x) = sin4(x) – cos3(x)?
Treat these as chain rules:
The derivative of sin4(x) is therefore: 4sin3(x)cos(x)
The derivative of cos3(x) is therefore: 3cos2(x) * –sin(x) or –3cos2(x)sin(x)
This gives us 4sin3(x)cos(x) + 3cos2(x)sin(x), which really cannot be simplified any further.
Example Question #12 : Other Differential Functions
What is the the first derivative of f(x) = 2tan(x)?
cot(x) * ln(2) * 2tan(x)
sec2(x) * ln(2) * 2tan(x)
sec(x) * tan(x) * ln(2) * 2tan(x)
ln(2) * 2tan(x)
ln(2) * 2sec(x)tan(x)
sec2(x) * ln(2) * 2tan(x)
This is a case of the chain rule.
Step 1: Deal with the exponential function
ln(2) * 2tan(x)
Step 2: Deal with the exponent
sec2(x)
Multiply them together to get your answer:
sec2(x) * ln(2) * 2tan(x)
Example Question #201 : Functions
What is the first derivative of f(x) = sin4(x) – cos4(x)?
4
4sin(4x)
1
2sin(2x)
4sin3(x) + 4cos3(x)
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 #15 : Other Differential Functions
What is the first derivative of f(x) = x2sin(4x3)?
2x * cos(4x3)
2x(sin(4x3) + 6x3cos(4x3))
sin(4x3) + 12x2cos(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 #16 : Other Differential Functions
What is the first derivative of f(x) = x4 – x * sin(x–5)?
4x3 – cos(x–5)
4x3 + cos(x–5)
4x3 – sin(x–5) + (5 * cos(x–5)/x5)
4x3 + sin(x–5) – (5 * cos(x–5)/x5)
None of the other answers
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)?
5(ln(2) + 1)
5(ln(2) + (1/2))
5ln(2)/2x
5/x
5/2x
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 #202 : Functions
What is the slope of the tangent line at x = 2 for f(x) = 6/x2?
1.5
3
–3
–12
–1.5
–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 #18 : Other Differential Functions
What is the first derivative of f(x) = 2ln(cos(x)sin(x))?
2cos2(x)sec(x)csc(x)
None of the other answers
2cos(2x)sec(x)csc(x)
cot(x)cos2(x)
2cot(x)cos2(x)
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 #1236 : Calculus
Take the derivative of
Derivative does not exist
![{y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/4656/gif.latex)
![{y}'=(7+x)^{(7+x)^2}[21+\ln(7+x)+x]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/4657/gif.latex)
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
![{y}'=y[2(7+x)\ln(7+x)+7+x]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/21141/gif.latex)
![{y}'=(7+x)^{(7+x)^2}[2(7+x)\ln(7+x)+7+x]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/21142/gif.latex)
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