differentiate using the first principle y = 5x^2 + 1?

Answer 1

y+dy=5(x+dx)^2 +1
y+dy=5(x² +2xdx +dx²) +1
y+dy=5x²+1 +5(2xdx+dx²)
dy=10 xdx as dx² is very small
dy/dx= 10x

Answer 2

Given: y = 5x^2 + 1 -------------->(1)

When x increases to ( x + D(x) ), consequently y increases to ( y + D(y) ). (D = Delta)

Therefore, y + D(y) = 5( x + D(x) )^2 + 1 ----------->(2)

On subtracting equation (1) from (2),

=> D(y) = 5(x + D(x))^2 - 5x^2
=> D(y) = 5x^2 +10xD(x) + 5(D(x))^2 -5x^2
=> D(y) = 10xD(x) + 5(D(x))^2

On dividing throughout by D(x) and taking limit such that D(x) tends to 0,(denoted by ldx0 below)

=> ldx0 (D(y)/D(x)) = ldx0 (10xD(x) + 5(D(x))^2)/(D(x))
=> dy/dx= ldx0 (10xD(x)/(D(x)) + 5(D(x))^2/(D(x)))
=> dy/dx = ldx0 (10x + 5D(x))
=> dy/dx = ldx0 (10x) + 5ldx0(D(x))
=> dy/dx = 10x.

Because for lim with Dx tend to 0, 10x is a constant while 5D(x) would become 0.

Hence d(5x^2 + 1)/dx = 10x.

Answer 3

Assuming this problem was taken from some textbook, what is the first principle stated therein?

In any case, chances are you are looking for: 10x+k

Answer 4

Let y=f(x)=5x^2 + 1

= lim [delta y/ delta x ]
delta x -->0

= lim { [f(x + delta x)] -f(x) }/ delta x
delta x -->0

= lim [ {5(x + delta x)^2 + 1} - 5x^2 - 1] / delta x
delta x -->0

= lim {5[x^2 + 2x(delta x) + (delta x)^2] +1} - 5x^2 -1/ delta x
delta x -->0

= lim [5x^2 + 10x(delta x) + (delta x)^2 + 1 -1 - 5x^2 ]/delta x
delta x -->0

= lim [10x(delta x) + (delta x)^2] / delta x
delta x -->0

= lim (delta x)(10 x + delta x) / (delta x)
delta x -->0

= lim 10x + delta x
delta x -->0

= 10x (Shown)

Hope that helps. :)
where lim = limit and --> tends to

Answer 5

y=10x

Answer 6

y =5x^2+1 -------eqn 1
y+Δy=5(x+Δx)^2+1 where Δx-->0 ------ eqn 2

by subtracting eqn 1 from 2 we get
Δy=5{(x+Δx)^2-x^2}
Δy=5{x^2 + 2*x*Δx + Δx^2 -x^2}
dividing by Δx on both sides
Δy/Δx=(10*x*Δx)/Δx +Δx^2/Δx
Δy/Δx=10*x +Δx
putting value of Δx i.e - ->0
dy/dx=10*x

thats it.(note "-->" is tends to)

Answer 7

Allahu akbar, everyting is deriviable from the the first priciples defined through prophet Mohhmmad in the holy Koran. (a million/?x)' = lim as ?x has a tendency to 0 of [(a million/?(x+?x) - a million/?x)/?x] = lim as ?x has a tendency to 0 of [(?x-?(x+?x))/?x /(?(x+?x)?x)] = lim as ?x has a tendency to 0 of [(?x-?(x+?x))/?x]/ lim as ?x has a tendency to 0 of [a million/?(x+?x)?x)] = a million/x lim [as ?x has a tendency to 0 of [(?x-?(x+?x))/?x] = -a million/x lim as ?x has a tendency to 0 of [(?(x+?x)-?x)/?x] = i will multiply both nominator and denominator through (?(x+?x)+ ?x) = -a million/x lim as ?x has a tendency to 0 of [(?(x+?x)-?x)(?(x+?x)+ ?x)/(?x (?(x+?x)+ ?x))] = -a million/x lim as ?x has a tendency to 0 of [(x+?x- x)/(?x (?(x+?x)+ ?x))] = -a million/x lim as ?x has a tendency to 0 of [?x/(?x a million/(?(x+?x)+ ?x))] = -a million/x lim as ?x has a tendency to 0 of [a million/(?(x+?x)+ ?x))] = -a million/x lim as ?x has a tendency to 0 of [a million/(?(x+?x)+ ?x))] -a million/x a million/(?x + ?x) = -a million/2 a million/?x³

Answer 8

y' = 10x
i guess...