Math Problem Statement
1. Find dy/dx for the given functions. 2. Find dv/dx for the given functions.
Solution
Let's go through the problems in the image one by one.
1. Find dxdy
-
y=cos3(4x)
- Differentiate using the chain rule:
dxdy=3cos2(4x)⋅(−sin(4x))⋅4=−12cos2(4x)sin(4x)
-
y=(sin(x)+1)4
- Apply the chain rule:
dxdy=4(sin(x)+1)3⋅cos(x)
-
y=7tan3(1−2x)
- Differentiate using the chain rule:
dxdy=7⋅3tan2(1−2x)⋅(−2)sec2(1−2x)
Simplifying:
dxdy=−42tan2(1−2x)sec2(1−2x)
-
y=sin3(4x3)
- Differentiate using the chain rule:
dxdy=3sin2(4x3)⋅cos(4x3)⋅12x2=36x2sin2(4x3)cos(4x3)
-
y=csc2(3x)
- Use the chain rule:
dxdy=2csc2(3x)⋅(−cot(3x))⋅3=−6csc2(3x)cot(3x)
-
y=(1−cos(3x))3
- Use the chain rule:
dxdy=3(1−cos(3x))2⋅sin(3x)⋅3=9(1−cos(3x))2sin(3x)
-
y=−3sec3(2x)
- Differentiate using the chain rule:
dxdy=−3⋅3sec2(2x)⋅sec(2x)⋅tan(2x)⋅2=−18sec3(2x)tan(2x)
-
y=−2cos5(3x)
- Use the chain rule:
dxdy=−2⋅5cos4(3x)⋅(−sin(3x))⋅3=30cos4(3x)sin(3x)
-
y=(1−tan2(x))
- This simplifies using the identity 1−tan2(x)=sec2(x), so:
dxdy=2sec2(x)⋅sec2(x)=2sec4(x)
2. Find dxdv
-
v=sin(3x)cos(2x)
- Use the product rule and trigonometric identities:
dxdv=cos(3x)⋅3cos(2x)−sin(3x)⋅2sin(2x)
-
v=sin3(2x+1)
- Use the chain rule:
dxdv=3sin2(2x+1)⋅cos(2x+1)⋅2=6sin2(2x+1)cos(2x+1)
-
v=2xcos(3x)
- Use the product rule:
dxdv=2cos(3x)−2x⋅3sin(3x)=2cos(3x)−6xsin(3x)
-
v=1−tan(3x)
- Use the chain rule:
dxdv=21−tan(3x)1⋅(−sec2(3x))⋅3=21−tan(3x)−3sec2(3x)
-
v=3+2cos(2x)
- Use the chain rule:
dxdv=23+2cos(2x)1⋅(−2sin(2x))⋅2=3+2cos(2x)−2sin(2x)
Would you like further explanations on any of these derivatives? Here are 5 related questions for deeper understanding:
- How can the chain rule be extended to multivariable functions?
- What is the product rule and when is it applied?
- How does the derivative of tan(x) result in sec2(x)?
- Why is dxdcsc(x)=−csc(x)cot(x)?
- How do you apply trigonometric identities to simplify complex derivatives?
Tip: Always verify the derivative of trigonometric functions with identities before solving, as it often simplifies the work.