From this post, one type of derivative is

$\lim_{\Delta x\rightarrow0^+}\frac{f(a+\Delta x)-f(a)}{\Delta x}\equiv\frac{df(x)}{dx}\big|_{a^+}$

To be consistent with my previous interpretation of $0^+$ in this post, $\Delta x \rightarrow 0^+$ means constraining $\Delta x$ to positive numbers.

Next, define $x$ and $a$ as variables for nonnegative real numbers, to avoid having a negative angle for the sine function. In this post, I find the derivative of $\sin x$ using the previous definition of a derivative. I also approximately follow the steps in reference [1].

Using $f(x) = \sin x$, the derivative is

$ \frac{d \sin(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{\sin(a+\Delta x) – \sin(a)}{\Delta x} $.

The next step is to rewrite the numerator with

$ \sin(a+\Delta x) = \sin a \cos \Delta x + \sin \Delta x \cos a $,

from a trigonometric identity that was derived in this post. The limit becomes

$ \frac{d \sin(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \frac{ \sin a \cos \Delta x + \sin \Delta x \cos a – \sin a}{\Delta x} $.

From the distributive property,

$ \frac{d \sin(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+} \bigg( \frac{ \sin a (\cos \Delta x – 1)}{\Delta x}  + \frac{\sin \Delta x \cos a }{\Delta x} \bigg) $.

From this property,

$ \frac{d \sin(x)}{dx} \big|_{a^+} = \lim_{\Delta x \rightarrow 0^+}  \frac{ \sin a (\cos \Delta x – 1)}{\Delta x}  + \lim_{\Delta x \rightarrow 0^+} \frac{\sin \Delta x \cos a }{\Delta x} $.

And from this property,

$ \frac{d \sin(x)}{dx} \big|_{a^+} = -\sin a \lim_{\Delta x \rightarrow 0^+}  \frac{ 1 – \cos \Delta x}{\Delta x}  + \cos a \lim_{\Delta x \rightarrow 0^+} \frac{\sin \Delta x  }{\Delta x} $.

From this post, it was shown that

$ \lim_{\Delta x \rightarrow 0^+} \frac{1 – \cos \Delta x}{\Delta x} = 0$.

And from this post, it is clear that

$\lim_{\Delta x \rightarrow 0^+} \frac{\sin \Delta x  }{\Delta x} = 1$.

Inserting these results,

$ \frac{d \sin(x)}{dx} \big|_{a^+} = \cos a$

for any nonnegative real number $a$.

If $a=x$, then

$\boxed{ \frac{d \sin(x)}{dx} \big|_{x^+} = \cos x }$

for any nonnegative real number $x$.


[1] Morris Kline. Calculus, An Intuitive and Physical Approach. Dover Publications, 1998.

[2] David V. Widder. Advanced Calculus. Dover 1989.

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