I tried a lot to solve this question, and every time i get 2, while the correct answer is 4, HELP to solve it?

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#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#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)#

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

Point 1 #->P_1->(x,y)=(2,-1)#

#color(blue)("Determine the gradient")#

If #y=ax^n+c# where c is a constant then

Then #dy/dx=nxx(ax^(n-1)) = nax^(n-1)#

The constant just 'goes away' ( cancels out - see part 1)

These days they write #f'(x)# instead of #dy/dx#

#f'(x)=2x" "............................Equation(2)#

So at point 1: # x=2#

So by substituting for #x# in #Equation(2)# we have:

#f'(x)="gradient"=2(x)=2(2)=4# as required
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#color(blue)("Determine the equation of the tangent")#

If some unknown point is #(x_("any"),y_("any"))# then

Gradient #=m=4=("change in y")/("change in x")=(y_("given")-y_("any"))/(x_("given")-x_("any"))#

#m=4=(-1-y)/(2-x)#

#4(2-x)=-1-y#

#8-4x=-1-y#

#y=4x-8-1#

#y=4x-9#