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f′(x)=limh→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}f′(x)=limh→0hf(x+h)−f(x)
ddxsin(x)=cos(x)\frac{d}{dx} \sin(x) = \cos(x)dxdsin(x)=cos(x)
sin(x+h)=sin(x)cos(h)+cos(x)sin(h)\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)sin(x+h)=sin(x)cos(h)+cos(x)sin(h)
limh→0sin(h)/h=1\lim_{h \to 0} \sin(h)/h = 1limh→0sin(h)/h=1
The limit is 000; it's the derivative of cos\coscos at x=0x=0x=0.
ddxcos(x)=−sin(x)\frac{d}{dx} \cos(x) = -\sin(x)dxdcos(x)=−sin(x)
ddxtan(x)=sec2(x)\frac{d}{dx} \tan(x) = \sec^2(x)dxdtan(x)=sec2(x)
ddxsec(x)=sec(x)tan(x)\frac{d}{dx} \sec(x) = \sec(x)\tan(x)dxdsec(x)=sec(x)tan(x)
When dydx=0\frac{dy}{dx} = 0dxdy=0
ddxsin(x)=cos(x)\frac{d}{dx} \sin(x) = \cos(x)dxdsin(x)=cos(x), ddxcos(x)=−sin(x)\frac{d}{dx} \cos(x) = -\sin(x)dxdcos(x)=−sin(x)
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