Calc2.58

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f(x)=\sin x\,

f'(x)=\lim _{{\Delta x\to 0}}{\frac  {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{{\Delta x\to 0}}{\frac  {\sin(x+\Delta x)-\sin x}{\Delta x}}

=\lim _{{\Delta x\to 0}}{\frac  {\sin x\cos \Delta x+\cos x\sin \Delta x-\sin x}{\Delta x}}

=\lim _{{\Delta x\to 0}}{\frac  {\sin x(\cos \Delta x-1)}{\Delta x}}+\lim _{{\Delta x\to 0}}{\frac  {\cos x\sin \Delta x}{\Delta x}}

To do these limits, we must recall simple limits given to us in most calculus texts.

\lim _{{x\to 0}}{\frac  {\sin x}{x}}=1

\lim _{{x\to 0}}{\frac  {1-\cos x}{x}}=0

Regrouping what we have so far, we can use these limits we already know to finish this problem.

f'(x)=\sin x\lim _{{\Delta x\to 0}}{\frac  {\cos \Delta x-1}{\Delta x}}+\cos x\lim _{{\Delta x\to 0}}{\frac  {\sin \Delta x}{\Delta x}}=(\sin x)(0)+(\cos x)(1)=\cos x

Thus, we have shown the derivative of \sin x is \cos x


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