How to find formula for the nth derivative of #f(x)=cos(ax+b)#?

Question

How to find  formula for the nth derivative of #f(x)=cos(ax+b)#?

Emily Ammerman · Accepted Answer

#d^n/(dx^n)f(x)= {(n = 2k -> (-1)^k a^(2k)cos(ax+b)),(n=2k+1->(-1)^(k+1) a^(2k+1)sin(ax+b)):}#

Making #e^(i(ax+b)) = cos(ax+b)+isin(ax+b)# we have

#f(x) = "Re"(e^(i(ax+b)))# then

#d^n/(dx^n)f(x) = "Re"(d^n/(dx^n)e^(i(ax+b))) = "Re"((ia)^n e^(i(ax+b)))#

now if #n=2k# we have

#"Re"((-1)^k a^(2k)cos(ax+b)) = (-1)^k a^(2k)cos(ax+b)#

and if #n=2k+1#we have

#"Re"(i(-1)^k a^(2k+1)(cos(ax+b)+isin(ax+b))) = -(-1)^k a^(2k+1)sin(ax+b)#

Finally

#d^n/(dx^n)f(x)= {(n = 2k -> (-1)^k a^(2k)cos(ax+b)),(n=2k+1->(-1)^(k+1) a^(2k+1)sin(ax+b)):}#

Scarlett Allison · Answer

#f^(n) (x)=a^ncos(ax+b+npi/2).# #f(x)=cos(ax+b)# # rArr f'(x)=-sin(ax+b)d/dx(ax+b)=-asin(ax+b).# We note that, #f'(x)=a^1cos(ax+b+pi/2).# #f'(x)=-asin(ax+b)# # rArr f''(x)=-a(cos(ax+b))*a=-a^2cos(ax+b).# Or, #f''(x)=a^2cos(ax+b+pi)=a^2cos(ax+b+2pi/2).# #rArr f'''(x)=-a^2(-sin(ax+b))*a=a^3sin(ax+b)=a^3cos(ax+b+3pi/2).# Generalising, #f^(n) (x)=a^ncos(ax+b+npi/2).#

Samantha Adkison · Answer

#d^n/dx^n (cos(ax+b)) = a^ncos(ax+b+(npi)/2)#

Note that:

#d/dx (cos(ax+b)) = -asin(ax+b) = acos(ax+b+pi/2)#

Now suppose that:

#d^n/dx^n (cos(ax+b)) = a^ncos(ax+b+(npi)/2)#

then:

#d^(n+1)/(dx^(n+1)) (cos(ax+b)) = d/dx a^ncos(ax+b+(npi)/2) = a^(n+1) cos(ax+b+((n+1)pi)/2)#

We proved the formula for #k=1# and we proved that if it is valid for #k=n# then it is also valid for #k=n+1#, So, by induction:

#d^n/dx^n (cos(ax+b)) = a^ncos(ax+b+(npi)/2)#