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

Answer 1

See Below.

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.

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Answer 2

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

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#
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Answer 3

To find the derivative of ( \sin(x^\circ) ) using the first principle, we need to recall that ( x ) should be in radians for differentiation. However, if you're specifically asked to find the derivative of ( \sin(x^\circ) ) using the first principle (which is essentially the limit definition of the derivative), you would proceed as follows:

Let ( f(x) = \sin(x^\circ) ).

Using the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Substitute ( f(x) = \sin(x^\circ) ):

[ f'(x) = \lim_{h \to 0} \frac{\sin((x + h)^\circ) - \sin(x^\circ)}{h} ]

Now, using the angle addition formula for sine, we have:

[ \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) ]

Let's denote ( a = x^\circ ) and ( b = h^\circ ):

[ \sin((x + h)^\circ) = \sin(x^\circ)\cos(h^\circ) + \cos(x^\circ)\sin(h^\circ) ]

Plug this back into the expression for ( f'(x) ):

[ f'(x) = \lim_{h \to 0} \frac{\sin(x^\circ)\cos(h^\circ) + \cos(x^\circ)\sin(h^\circ) - \sin(x^\circ)}{h} ]

[ = \lim_{h \to 0} \frac{\sin(x^\circ)(\cos(h^\circ) - 1) + \cos(x^\circ)\sin(h^\circ)}{h} ]

Now, as ( h \to 0 ), ( \sin(h^\circ) ) approaches ( 0 ), and ( \cos(h^\circ) - 1 ) approaches ( 0 ). So, we have:

[ f'(x) = \lim_{h \to 0} \frac{\sin(x^\circ)(0) + \cos(x^\circ)(0)}{h} ]

[ = \lim_{h \to 0} 0 = 0 ]

Therefore, the derivative of ( \sin(x^\circ) ) using the first principle is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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