Question

Find The Derivative of Sinx° (degree) By Using The First Principal ?

Answer 1

See Below.

Explanation:

The First Principle Of Differentiation is :-

`d/dxf(x) = lim_(hrarr0) (f(x + h) - f(x))/(h)` where `h = dx`. [Small change in `x`].

So, Using it,

`d/dx(sin x) = lim _(h rarr 0) (sin(x + h) - sin(x))/h`

`= lim_(hrarr 0)(2cos(((x + h) + h)/2)sin((x + cancelh cancel(- h))/2))/h`

[Using `sin A - sin B = 2cos((A + B)/2)sin((A - B)/2)`}]

`= lim_(hrarr 0)(2cos((2x + h)/2)sin(h/2))/h`

`= lim_(hrarr0)(cos(x + h/2))sin(h/2)/(h/2)`

We know, `lim_(xrarr0) (sin x)/x = 1`.

So,

The Limit

`= lim _(hrarr0) cos(x + h/2) * 1` [As `h/2 rarr 0` when `h rarr 0`]

`= lim_(hrarr 0) cos (x + h/2)`

`= cos(x + 0/2) = cos x`.

Hope this helps.

Answer 2

`f'(x)=pi/180cos((pix)/180)^R`
or
`f'(x)=pi/180cosx^circ`

Explanation:

We know that,

`x^circ=((pix)/180)^R=(pix)/180`

Let,

`f(x)=sin((pix)/180)=>f(t)=sin((pit)/180)`

Now,

`color(blue)(f'(x)=lim_(t tox) (f(t)-f(x))/(t-x)`

`color(white)(f'(x))=lim_(t tox)(sin((pit)/180)-sin((pix)/180))/(t-x)`

`color(white)(f'(x))=lim_(t tox)(2cos(((pit)/180+(pix)/180)/2)sin(((pit)/180-(pix)/180)/2))/(t-x)`

`color(white)(f'(x))=lim_(t tox)(2cos(pi/360(t+x))sin((pi/360(t-x))))/((pi/360(t-x)))*pi/360`

`color(white)(f'(x))=(2pi)/360lim_(t tox)cos(pi/360(t+x))*lim_(t tox)[sin(pi/360(t-x))/(pi/360(t-x))]`

Now,

`t tox=>(t-x)to0=>pi/360(t-x)to0 and lim_(thetato0)sintheta/theta`=`1`

`:.f'(x)=pi/180cos(pi/360(x+x))*(1)`

`f'(x)=pi/180cos(pi/360(2x))`

`f'(x)=pi/180cos((pix)/180)^R` , where ,`(pix)/180=x^circ`

`f'(x)=pi/180cosx^circ`