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#