I need help finding the derivative of 2sin(x) because I keep on getting 2sin(x)cos(x) instead of 2cos(x). Do I use the chain rule?
The Correct Answer and Explanation is:
You’re on the right track, but it looks like you might be confusing the result for the derivative of sin(2x)\sin(2x)sin(2x) with the derivative of 2sin(x)2\sin(x)2sin(x). Let’s clarify the steps.
To find the derivative of 2sin(x)2\sin(x)2sin(x), you do not need the chain rule because there is no composite function like sin(2x)\sin(2x)sin(2x) involved. Instead, you just apply basic differentiation rules.
Derivative of 2sin(x)2\sin(x)2sin(x)
- Constant Multiplier Rule: The constant multiple (in this case, 2) can be factored out of the derivative operation. So, you are left with: ddx[2sin(x)]=2⋅ddx[sin(x)]\frac{d}{dx}[2\sin(x)] = 2 \cdot \frac{d}{dx}[\sin(x)]dxd[2sin(x)]=2⋅dxd[sin(x)]
- Derivative of sin(x)\sin(x)sin(x): The derivative of sin(x)\sin(x)sin(x) with respect to xxx is simply cos(x)\cos(x)cos(x). Therefore: 2⋅cos(x)2 \cdot \cos(x)2⋅cos(x)
Thus, the derivative of 2sin(x)2\sin(x)2sin(x) is 2cos(x)2\cos(x)2cos(x).
Why not 2sin(x)cos(x)2\sin(x)\cos(x)2sin(x)cos(x)?
The expression 2sin(x)cos(x)2\sin(x)\cos(x)2sin(x)cos(x) is the result you would get if you were differentiating sin(2x)\sin(2x)sin(2x) using the chain rule. Here’s why:
When you differentiate sin(2x)\sin(2x)sin(2x), the chain rule tells you to first differentiate the outer function sin(u)\sin(u)sin(u), where u=2xu = 2xu=2x, and then multiply by the derivative of 2x2x2x. This process leads to the following:ddx[sin(2x)]=cos(2x)⋅ddx[2x]=2cos(2x)\frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot \frac{d}{dx}[2x] = 2\cos(2x)dxd[sin(2x)]=cos(2x)⋅dxd[2x]=2cos(2x)
But that’s not the case here. Since you’re differentiating 2sin(x)2\sin(x)2sin(x), you only need to apply the basic rule for the derivative of sin(x)\sin(x)sin(x), resulting in 2cos(x)2\cos(x)2cos(x).
Conclusion
The correct derivative of 2sin(x)2\sin(x)2sin(x) is 2cos(x)2\cos(x)2cos(x), and no chain rule is needed. You only use the chain rule when differentiating composite functions.
