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 sinx is cosx


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