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

2 Answers
Jun 16, 2018

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.

Jun 16, 2018

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