Limit as Derivative Basis

Limit as Derivative Basis
The concept of limit is the approximate value. Logically the approximation value is a value that is not true but can be completely. The derivative of the function comes from the logical concept of a limit that approaches the value 0. The following is the definition of the derivative of the function:
$$f'(x)=\text{lim}_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

From the above definition we can find $f'(x)$ from the function $f(x)=ax^n$ as follows:
$$f(x)=ax^n$$ $$f'(x)=\text{lim}_{h \to 0} \frac{a(x+h)^n-ax^n}{h}$$ $$f'(x)=\text{lim}_{h \to 0} \frac{ax^n+a.n.h.x^{n-1}+... -ax^n}{h}$$ $$f'(x)=\text{lim}_{h \to 0} a.n.x^{n-1}=a.n.x^{n-1}$$
We will try to find $D_x~\text{sin }x$, yakni kita peroleh bentuk: $$\frac{\text{sin}(x+h)-\text{sin}x}{h}$$ $$=\text{sin}x.\frac{(\text{cos}h-1)}{h}+\frac{\text{sin}h}{h}.\text{cos}x$$ $$=\text{sin}x.m+n.\text{cos}x$$ We can find the limit values ​​of $m$ and $n$ by using the approximate value of $h=0.0000001$ (meaning $h$ is close to 0), we will get $m=0$ and $n=1$. So $D_x~\text{sin}x=\text{cos}x$.
Thanks for view, see you and happy learning.

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