Question

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

Answer 1

`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)):}`

Explanation:

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)):}`

Answer 2

`f^(n) (x)=a^ncos(ax+b+npi/2).`

Explanation:

`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).`

Answer 3

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

Explanation:

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)`