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mherna48
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Hey can anyone help me prove that the derivative of sin(x)/x is zero at x=0
`snipez90 said:A function f:R -> R is differentiable at 0 if
[tex]\lim_{h \rightarrow 0}\frac{f(h)-f(0)}{h}[/tex]
exists. You can apply this definition and then use l'Hopital's rule or perhaps some first-order estimates to show that this last limit exists. Alternatively, you can probably work directly with the series expansion for sin.
The derivative of sin(x)/x is given by the quotient rule as (x(cos(x)) - sin(x))/x^2.
To prove that the derivative of sin(x)/x is zero at x=0, we need to evaluate the derivative at x=0 and show that it equals zero. This can be done by using the limit definition of a derivative and simplifying the expression until we get 0 as the result.
This proof is important because it is a fundamental result in calculus and helps us understand the behavior of functions near a point. It also allows us to use this result in further calculations and applications.
The intuition behind this result is that as x approaches 0, the function sin(x)/x becomes more and more similar to the function sin(x), which has a derivative of cos(x). However, at x=0, the function is undefined, so we need to use the limit definition to evaluate the derivative, and we end up with 0 as the result.
Yes, there are other proofs available for this result, such as using the power series expansion of sin(x) and the quotient rule, or using the chain rule and the fact that sin(x) is an odd function. However, the most common and straightforward proof is using the limit definition of a derivative.