How do you use the definition of a derivative to find the derivative of #f(x)=cosx#?

Question

Evelyn Agard · Accepted Answer

#d/(dx) cosx=-sinx#

By definition:

#f'(x) = lim_(Deltax->0) (f(x+Deltax)-f(x))/(Deltax)#

In our case, #f(x) = cosx#, so:

#d/(dx) cosx= lim_(Deltax->0) (cos(x+Deltax)-cos(x))/(Deltax)#

We can now use the trigonometric identity:

#cos(x+Deltax) = cosxcos(Deltax)-sinxsin(Deltax)#

and obtain:

#d/(dx) cosx=lim_(Deltax->0)(cosxcos(Deltax)-sinxsin(Deltax)-cosx)/(Deltax)#

that is:

#d/(dx) cosx=lim_(Deltax->0)cosx(1-cos(Deltax))/(Deltax)-sinx(sin(Deltax))/(Deltax)#

But:

#lim_(t->0) sint/t=1#

and:

#lim_(t->0) (1-cost)/t=0#

and we can conclude:

#d/(dx) cosx=-sinx#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = \cos(x) $ using the definition of a derivative, you apply the limit definition of the derivative, which states:

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

For $ f(x) = \cos(x) $, substitute $ \cos(x) $ into the definition:

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

Using the trigonometric identity $ \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) $, we rewrite the numerator:

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

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

$$ = \lim_{h \to 0} \frac{-2\sin^2(h/2)\sin(x)}{h} $$

Now, using the trigonometric limit $ \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1 $, we can evaluate the limit:

$$ f'(x) = -\sin(x) \lim_{h \to 0} \frac{2\sin(h/2)}{h/2} $$

$$ = -\sin(x) \cdot 2 $$

$$ = -2\sin(x) $$

So, the derivative of $ f(x) = \cos(x) $ is $ f'(x) = -2\sin(x) $.