Find the derivative of #sin^2x# using first principles?

Answer 1

See below

It is tedious (and space-consuming) to use #lim_(hrarr0)# on every line, so I hope you'll excuse my approach. We'll simplify the difference quotient first, then find the limit.
# ((sin(x+h))^2-(sinx)^2)/h = ((sinxcos h +cos x sin h)^2-(sinx)^2)/h#
# = (color(red)(sin^2xcos^2h)+2sinxcos h cosx sin h +color(blue)(cos^2xsin^2h)-color(red)(sin^2x))/h#
# = (sin^2x(cos^2h-1))/h +(2sinxcos h cosx sin h)/h +(cos^2xsin^2h)/h#
# = sin^2x(cosh-1)/h(cosh+1) +2sinxcos h cosx sin h /h+cos^2xsin h/h sin h#
Taking limit as #h rarr0#, we get
#sin^2(0)(0)(cos 0 +1) +2sinx (cos 0) cosx (1) +cos^0 (1) sin0#
# = 0+2sinx cos x +0#
# = 2sinx cos x#
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Answer 2

# f'(x) =2sinxcosx #

The definition of the derivative of #y=f(x)# is
# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #
So with # f(x) = sin^2x # then;
# f'(x)=lim_(h rarr 0) {sin^2(x+h) -sin^2(x)}/h#
Let us focus on the numerator #f(x+h)-f(x)#;
# f(x+h)-f(x) # # \ \ = sin^2(x+h) -sin^2(x)# # \ \ = (sinxcos h+cosxsin h)^2 -sin^2(x)# # \ \ = (sinxcos h)^2+2sinxcos hcosxsin h + (cosxsin h)^2 -sin^2(x)# # \ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x# # \ \ = sin^2xcos^2h+2sinxcos hcosxsin h + cos^2xsin^2h -sin^2x# # \ \ = sin^2x(cos^2h-1)+2sinxcos hcosxsin h + cos^2xsin^2h # # \ \ = sin^2xsin^2h+2sinxcos hcosxsin h + cos^2xsin^2h # # \ \ = sin^2h(sin^2x+cos^2x) + 2sinxcos hcosxsin h # # \ \ = sin^2h+ 2sinxcos hcosxsin h #

And so the limit becomes:

# f'(x)=lim_(h rarr 0) {sin^2h+ 2sinxcos hcosxsin h}/h# # " "=lim_(h rarr 0) {sin^2h/h+ (2sinxcos hcosxsin h)/h}# # " "=lim_(h rarr 0) {sin h * sin h/h+ 2sinxcosx*cos h*sin h/h}#

And then using the Fundamental trigonometric calculus limits:

# lim_(theta rarr0) sin theta /theta = 1 #

we have:

# f'(x)= 0 * 1+ 2sinxcosx*1*1# # " "=2sinxcosx #
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Answer 3

To find the derivative of (\sin^2(x)) using first principles, we start with the definition of the derivative:

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

For (f(x) = \sin^2(x)), we have:

[ f(x) = \sin^2(x) ] [ f(x + h) = \sin^2(x + h) ]

Now, let's substitute these into the definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{\sin^2(x + h) - \sin^2(x)}{h} ]

Now, we can use the identity (\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)):

[ \sin^2(x + h) = (\sin(x + h))^2 = (\sin(x)\cos(h) + \cos(x)\sin(h))^2 ]

[ = \sin^2(x)\cos^2(h) + 2\sin(x)\cos(x)\sin(h)\cos(h) + \cos^2(x)\sin^2(h) ]

[ - \sin^2(x) ]

Now, substitute this expression back into the derivative:

[ f'(x) = \lim_{h \to 0} \frac{\sin^2(x)\cos^2(h) + 2\sin(x)\cos(x)\sin(h)\cos(h) + \cos^2(x)\sin^2(h) - \sin^2(x)}{h} ]

[ = \lim_{h \to 0} \frac{\sin^2(x)\cos^2(h) + 2\sin(x)\cos(x)\sin(h)\cos(h) + \cos^2(x)\sin^2(h) - \sin^2(x)}{h} ]

[ = \lim_{h \to 0} \frac{\sin^2(x)(\cos^2(h) - 1) + 2\sin(x)\cos(x)\sin(h)\cos(h) + \cos^2(x)(\sin^2(h))}{h} ]

[ = \lim_{h \to 0} \frac{\sin^2(x)(\cos^2(h) - 1)}{h} + \lim_{h \to 0} \frac{2\sin(x)\cos(x)\sin(h)\cos(h)}{h} + \lim_{h \to 0} \frac{\cos^2(x)(\sin^2(h))}{h} ]

Now, as (h) approaches 0, (\cos^2(h) - 1) approaches 0, (\sin(h)) approaches 0, and (\sin^2(h)) approaches 0. So:

[ f'(x) = 0 + 0 + 0 = 0 ]

Therefore, the derivative of (\sin^2(x)) using first principles 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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