The problem for session 4: How can I calculate f'(x) of f(x)=sin2x as 2cos2x? Have I missed anything?

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The problem for session 4: How can I calculate f'(x) of f(x)=sin2x as 2cos2x? Have I missed anything?

MIT 18.01 Single Variable Calculus (OCW)
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This uses the chain rule. the derivative of sin x is cos x, but we do have sin x. We have sin 2x.
Taking the derivative will give us cos 2x times the derivative of 2x. Since the derivative of 2x is 2, the answer is (cos 2x)*2, or 2*cos 2x
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You can substitute 2x with u. Differentiate sinu. Then differentiate 2x. Multiply both of them which as has been previously mentioned is an application of the chain rule. Dont forget to replace u with 2x in your final answer.
After I learned Session 7, I knew that the derivative of sinx is cosx. If I replace 2x with u and calculate again, wouldn't it be f'(1/2u)=cosu, and f'(x)=cos2x? Where did preceding 2 come from? What I have calculated is, using sin2x=2sinxcosx, |dw:1328108260240:dw| What is wrong?
Hi makopo Via the Chain Rule (also called the Substitution Rule): \[y =\sin 2x\] Let \[u = 2x\] So, \[dy/dx = dy/du \times du/dx\] (Chain/Substitution Rule) \[dy/dx = d/du (\sin u) \times d/dx (2x)\] Solving this, \[dy/dx = \cos u \times 2\] Substituting for u (given above) \[dy/dx = 2\cos 2x, u = 2x\] Hope this was helpful!
I took Session11 just ago, learned chain rule, after then returned to Session4 and challenged again. This time I can understand why sin2x = 2cos2x! The problem is that this problem is posted on Session 4, though you can't solve it before taking Session 11. I'll send a feedback about this issue... Thank you all, especially DaveJohnson, for noticing me that I need to proceed with some sessions to know the "chain rule" to solve this problem.
The way you rewrote sin 2x as 2 sin x cos x means we now have a product. Taking the derivative of a product requires the use of the product rule.
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