Find the derivative of #cos# using First Principles?

Question

Cameron Alexander · Accepted Answer

By the limit definition of the derivative if #y=f(x)#, then # dy/dx=f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h # So with # y=f(x) = cosx # we have; # f'(x) = lim_(h rarr 0) ( cos(x+h) - cos x ) / h # Using the cosine sum of angle formula: # cos(A+B)=cosAcosB- sinAcosB # We get # f'(x) = lim_(h rarr 0) ( cosxcos h-sinxsin h - cos x ) / h #
# "     " = lim_(h rarr 0) ( cosxcos h-cosx-sinxsin h ) / h #
# "     " = lim_(h rarr 0) ( cosx(cos h-1)-sinxsin h ) / h #
# "     " = lim_(h rarr 0) {(cosx(cos h-1))/h-(sinxsin h ) / h} #
# "     " = lim_(h rarr 0) (cosx(cos h-1))/h- lim_(h rarr 0)(sinxsin h ) / h #
# "     " = cosxlim_(h rarr 0) (cos h-1)/h - sinx lim_(h rarr 0)(sin h ) / h # We now have to rely on some standard calculus limits: # lim_(h rarr 0)sin h/h =1 #
# lim_(h rarr 0)(cos h-1)/h =0 # And so using these we have: # f'(x) = cosx(0) - sinx (1) #
# "     " = - sinx # Hence, # dy/dx=-sinx#

Ezekiel Alexander · Answer

undefined To find the derivative of $ \cos(x) $ using first principles:

1. Start with the limit definition of the derivative:

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

2. Substitute $ f(x) = \cos(x) $ into the limit definition:

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

3. Use the trigonometric identity $ \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) $:

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

4. Expand and simplify the expression:

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

5. Apply the limit laws and trigonometric limit properties to evaluate the limit:

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

6. Recognize that $ \lim_{h \to 0} \frac{\sin(h)}{h} = 1 $ and $ \lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0 $:

$$ f'(x) = -\sin(x) \cdot 1 + \cos(x) \cdot 0 $$
$$ = -\sin(x) $$

Therefore, the derivative of $ \cos(x) $ with respect to $ x $ is $ -\sin(x) $.