Question

Differentiate `sin(x/2)` using first principles?

Answer 1

`d/dx sin(x/2)=1/2cos(x/2)`

Explanation:

Let:

`f(x) = sin(x/2)`

By definition, the derivative of `f(x)` is the limit:

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So for the given function, we have;

`f'(x) = lim_(h rarr 0) ( sin((x+h)/2) - sin (x/2) ) / h`
`" " = lim_(h rarr 0) ( sin(x/2+h/2) - sin (x/2) ) / h`

Applying the trigonometric identity:

`sin (A+B)=sinAcosB+sinBcosA`

We get

`f'(x ) =lim_(h rarr 0) ( sin(x/2)cos(h/2)+sin(h/2)cos(x/2) - sin(x/2) ) / h`
`" " = lim_(h rarr 0) ( sin(x/2)(cos (h/2)-1)+sin(h/2)cos(x/2) ) / h`

`" " = lim_(h rarr 0) ( (sin(x/2)(cos (h/2)-1))/h+(sin(h/2)cos(x/2)) / h )`

`" " = lim_(h rarr 0) (sin(x/2)(cos(h/2)-1))/h+lim_(h rarr 0)(sin (h/2)cos(x/2)) / h`

`" " = sin(x/2) \ lim_(h rarr 0) (cos (h/2)-1)/h + cos(x/2) \ lim_(h rarr 0)(sin (h/2)) / h`

We know have to rely on some standard limits:

`lim_(theta rarr 0)sin theta/theta =1`
`lim_(h rarr 0)(cos theta-1)/h =0`

And so manipulating the denominator by a factor of `2` to match the sine and cosine variable we have:

`f'(x) = sin(x/2) \ lim_(h rarr 0) (1/2)(cos (h/2)-1)/(h/2) + cos(x/2) \ lim_(h rarr 0)((1/2)sin (h/2)) / (h/2)`
`" " = 1/2sin(x/2) \ lim_(h rarr 0) (cos (h/2)-1)/(h/2) + 1/2cos(x/2) \ lim_(h rarr 0)(sin (h/2)) / (h/2)`

And putting `theta=h/2` and noting that `h/2 rarr 0` as `h rarr 0` we have:

`f'(x) = 1/2sin(x/2) \ lim_(theta rarr 0) (cos theta-1)/(theta) + 1/2cos(x/2) \ lim_(theta rarr 0)(sin theta) / (theta)`

`" " = 1/2sin(x/2) xx 0 + 1/2cos(x/2) xx 1`
`" " = 1/2cos(x/2)`

Hence,

`d/dx sin(x/2)=1/2cos(x/2)`