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# Proof of the derivative of sin(x)

Let's dive into the proof that the derivative of sin(x) equals cos(x). By applying angle addition formulas, the squeeze theorem, and exploring the concept of limits, we unravel this interesting proof. This exploration not only solidifies our understanding, but also equips us to handle more complex derivatives in the future.

## Want to join the conversation?

• At , Sal finished writing a very long expression:
lim ∆x->0 [(cos x sin∆x + sin x cos ∆x - sin x)/x]
I tried evaluating and got a wrong answer that the whole limit =(sinx-sinx)/x= 0/x, but why can't I just evaluate the whole thing here instead of using the limit properties and go through a lot of steps to get the final answer? Why is it wrong to do that? Does it have something to do with the limit properties?
• If you just evaluate it without using properties, you'll get 0/0 which is not possible. Therefore, you need to use the properties to get rid of this problem. Also, if you actually use a graphing tool to draw it, and find the limit, you'll get the value that you got when you evaluate it.
• how can you change sin(x+ delta x)into cosx*sin(delta x)+sinx*cos(delta x)
how can i find the formula?
• Limit of sin(1÷x) as x approaches to 0
• How does it make sense what Sal did on to take out the cosx like that
• One of the properties of limits is that the limit of f(x)*g(x) = limit of f(x) * limit of g(x). Sal applied this rule to transform the original limit into the product of the limits of cos(x) and sin(Δx)/Δx. Since cos(x) does not change with respect to Δx, the limit of cos(x) is simply cos(x). This left us with cos(x) * limit of sin(Δx)/Δx.