How do you find the derivative of #(cos x)# using the limit definition?

1 Answer
Jim H
May 7, 2016

See the explanation section below.

Explanation:

We'll need the following facts:

From trigonometry:
#cos(A+B) = cosAcosB-sinAsinB#

Fundamental trigonometric limits:

#lim_(theta rarr0) sin theta /theta = 1#

#lim_(theta rarr0) (cos theta - 1) /theta = 0#

And here we go:

#f(x) = cosx#

#f'(x) = lim_(hrarr0)(cos(x+h)-cosx)/h#

# = lim_(hrarr0)(cosxcosh-sinxsinh-cosx)/h#

# = lim_(hrarr0)(cosxcosh-cosx-sinxsinh)/h#

# = lim_(hrarr0)((cosxcosh-cosx)/h-(sinxsinh)/h)#

# = lim_(hrarr0)(cosx(cosh-1)/h-sinx(sinh)/h)#

# = cosx(lim_(hrarr0)(cosh-1)/h)-sinx(lim_(hrarr0)(sinh)/h)#

# = cosx(0)-sinx(1)#

# = -sinx#

Related questions
See all questions in Limit Definition of Derivative
Impact of this question
41075 views around the world
You can reuse this answer
Creative Commons License