Using the definition of derivative, how do you prove that (cos⁡x)′=−sin⁡x?

Question

Deandre Barron · Accepted Answer

Remembering the cosine difference-to-product formula, that says:cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)and the fundamental limit:limx→0sin⁡xx=1, or in the most general writing:limf(x)→0sin⁡f(x)f(x)=1than:f′(x)=limh→0f(x+h)−f(x)h=limh→0cos⁡(x+h)−cos⁡xh==limx→0−2sin⁡(x+h+x2)sin⁡(x+h−x2)h==limh→0−2sin⁡(x+h2)sin⁡(h2)h==limh→0−2sin⁡(x+h2)sin⁡(h2)h2==limh→0−sin⁡(x+h2)⋅limh→0sin⁡(h2)h2=−sin⁡x⋅1=−sin⁡x

Dexter Conner · Accepted Answer

Use the formula for the cosine of a sum:

\cos (a + b) = \cos a \cos b - \sin a \sin b

together with the fundamental tigonometric limits:

lim_{h \to 0} \frac{\sin x}{x} = 1

and

lim_{h \to 0} \frac{\cos x - 1}{x} = 0

For

f (x) = \cos x

, we get:

f^{'} (x) = lim_{h \to 0} \frac{\cos (x + h) - \cos x}{h}

= lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}

= lim_{h \to 0} \frac{\cos x \cos h - \cos x - \sin x \sin h}{h}

= lim_{h \to 0} (\cos x \frac{\cos h - 1}{h} - \sin x \frac{\sin h}{h})

= lim_{h \to 0} \cos x lim_{h \to 0} \frac{\cos h - 1}{h} - lim_{h \to 0} \sin x lim_{h \to 0} \frac{\sin h}{h}

= (\cos x) (0) - (\sin x) (1) = - \sin x

Using the definition of derivative, how do you prove that

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