Given: #y=x^2-5 " ".......................Equation(1)#

Required: slope and tangent at point #P->(x,y)=(2,-1)#

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#color(blue)("First principle method")#

I am a bit 'old fashioned so I use #dy/dx# for differentiation instead of #f'(x) larr# f prime of x

Let #x# increase by the minute amount #deltax#

As we have change #x# then #y# will also change.

Let the small change in #y# be #deltay#

Then with the change #Equation(1)# becomes:

#y+deltay=(x+deltax)^2-5#

Expanding the brackets it becomes:

#y+deltay=x^2+2xdeltax+(deltax)^2" ".....................Equation(2)#

#Eqn(2)-Eqn(1)#

#y+deltay=x^2+2xdeltax+(deltax)^2#

#ul( ycolor(white)("dddd") = x^2color(white)("dddddddddddd")larr" Subtract")#

#0+deltay= 0+2xdeltax+(deltax)^2#

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#deltay= 2xdeltax+(deltax)^2" ".............................Equation(3)#

We need the rate of change #->("Change in "y)/("Change in "x) ->(deltay)/(deltax)#

So to have this on the left in #Eqn(3)# divide both sides by #deltax#

#(deltay)/(deltax)= (2xcancel(deltax))/(cancel(deltax))+(deltax)^(cancel(color(white)(.)2))/(cancel(deltax))#

#(deltay)/(deltax)= color(white)("d")2xcolor(white)("d.")+color(white)("..")deltax" ".........................Equation(3_a)#

Consider the priciple that #100/50=50/25=25/(12 1/2) =2#

As the values become less and less but still retain the same ratio it does not change the actual answer when dividing the denominator into the numerator. So as the value of #deltay and deltax# become less and less we still have the same intrinsic value. Consequently we can and may take each one very close to 0.

Limit as #x# approaches 0 is written as #lim_(x->0)#

#lim_(x->0)(deltax)/(deltay)color(white)("d") =color(white)("d")dy/dx color(white)("d")=ubrace(lim_(x->0)2x)+ubrace(lim_(x->0)deltax)#

#color(white)("dddddddddddd")dy/dxcolor(white)("d")=color(white)("ddd")2xcolor(white)("d")+color(white)("d")0#

#color(white)("dddddddddddd")color(brown)(dy/dxcolor(white)("d")=color(white)("d")2x)#

The modern way of writing this is:

if #f(x)=x^2-5 color(white)("d")# then #color(blue)(color(white)("d")dy/dx =f'(x)=2x)#