Differentiate #xsinx# using first principles?

Answer 1

#f'(x)=xcosx+sinx#

Derivation from first principles tells us that for a function #f(x)#, #f'(x)=lim_(h->0)(f(x+h)-f(x))/h#
In this case, #f(x)=xsinx#, so we have: #f'(x)=lim_(h->0)((x+h)sin(x+h)-xsinx)/h#
We can use the identity #sin(A+B)=sinAcosB+sinBcosA#
#color(white)(f'(x))=lim_(h->0)((x+h)(sin(x)cos(h)+cos(x)sin(h))-xsinx)/h#
#color(white)(f'(x))=lim_(h->0)(xsin(x)cos(h)+xcos(x)sin(h)+hsin(x)cos(h)+hcos(x)sin(h)-xsinx)/h#
#color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+(xcos(x)sin(h))/h+(hsin(x)cos(h))/h+(hcos(x)sin(h))/h-(xsinx)/h)#
#color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+(xcos(x)sin(h))/h+sin(x)cos(h)+cos(x)sin(h)-(xsinx)/h)#
It is known that: #lim_(h->0)sinh/h=1#
#color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+xcos(x)+sin(x)cos(h)+cos(x)sin(h)-(xsinx)/h)#
#color(white)(f'(x))=xcos(x)+sin(x)cos(0)+cos(x)sin(0)+lim_(h->0)((xsin(x)cos(h))/h-(xsinx)/h)#
#color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)cos(h))/h-(xsinx)/h)#
#color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)cos(h)-xsinx)/h)#
#color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)(cos(h)-1))/h)#
#color(white)(f'(x))=xcos(x)+sin(x)+xsin(x)lim_(h->0)(cos(h)-1)/h#
It is also known that: #lim_(h->0)(cos(h)-1)/h=0#
#color(white)(f'(x))=xcos(x)+sin(x)+xsin(x)0#
#color(white)(f'(x))=xcos(x)+sin(x)#
Proof: #f(x)=xsinx#
#f'(x)=d/dx[x]sinx+xd/dx[sinx]# #color(white)(f'(x))=1sinx+xcosx# #color(white)(f'(x))=xcosx+sinx#
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Answer 2

# d/dx xsinx = x cosx +sinx #

Let us define:

# f(x) = xsinx#

Using the limit definition of the derivative, we compute the derivative using:

# f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h #
# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ((x+h)sin(x+h) - xsinx)/h #
# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) {x((sin(x+h) - sinx))/h +sin(x+h)} #
# \ \ \ \ \ \ \ \ \ = x lim_(h rarr 0) ((sin(x+h) - sinx))/h +lim_(h rarr 0)sin(x+h) #
# \ \ \ \ \ \ \ \ \ = x {lim_(h rarr 0) (sin(x+h) - sinx)/h} +sinx #

So we only need to calculate the limit:

# L = lim_(h rarr 0) (sin(x+h) - sinx)/h #
Which, coincidentally, is the limit arising from the derivative of #sinx# (by definition). Now using the trigonometric multiply angle identity:
# sin(A+B) -= sinA cosB+cosA sinB #

We get:

# L = lim_(h rarr 0) (sin x cos h + cos x sin h - sin x)/h #
# \ \ \ = lim_(h rarr 0) (sin x (cos h -1))/h+ (cos x sin h)/h #
# \ \ \ = sinx \ {lim_(h rarr 0) (cos h -1)/h } + cos x \ {lim_(h rarr 0) ( sin h)/h} #
Both of these limits are standard calculus limits, and providing #x# is in radians, we have:
# lim_(h rarr 0) (cos h -1)/h =0 \ \ # and # \ \ lim_(h rarr 0) ( sin h)/h = 1 #

With these results, we get the result:

# L = cos x => f'(x) = x cosx +sinx #
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Answer 3

To differentiate ( x\sin(x) ) using first principles:

  1. Start with the definition of the derivative: [ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

  2. Substitute ( f(x) = x\sin(x) ) into the definition: [ f'(x) = \lim_{h \to 0} \frac{(x + h)\sin(x + h) - x\sin(x)}{h} ]

  3. Expand and simplify the expression: [ f'(x) = \lim_{h \to 0} \frac{x\sin(x) + h\sin(x) + x\sin(h) + h\sin(h) - x\sin(x)}{h} ] [ f'(x) = \lim_{h \to 0} \frac{h\sin(x) + x\sin(h) + h\sin(h)}{h} ] [ f'(x) = \lim_{h \to 0} \left(\sin(x) + \frac{x\sin(h) + h\sin(h)}{h}\right) ]

  4. Use the limit properties to separate the limit: [ f'(x) = \sin(x) + \lim_{h \to 0} \left(\frac{x\sin(h) + h\sin(h)}{h}\right) ]

  5. Recognize that ( \lim_{h \to 0} \frac{x\sin(h)}{h} = x ) and ( \lim_{h \to 0} \frac{h\sin(h)}{h} = 0 ): [ f'(x) = \sin(x) + x ]

Therefore, the derivative of ( x\sin(x) ) with respect to ( x ) using first principles is ( \sin(x) + x ).

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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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