Approximation using derivatives of cube root and power. Teach me.?

a) cube root of 70.

∛70

Let y= ∛x

→ dy/dx= (1/3)/(x) ^(2/3)

Let 70= x+∆x

i.e ∆x = 6

And x= 64 a number of whose cube root is known

y= ∛x and y+∆y= ∛(x+∆x)

By linear approximation ∆y= (dy/dx)∆x

Let y=∛64 = 4

dy/dx= (1/3)/(x) ^(2/3)= = (1/3)/(64) ^(2/3) =1/48

∆y= (dy/dx)∆x =∆y= (1/48)×6 =1/8=0.125

Hence

∛70 =y+∆y=4+0.15= 4.125

and y

b) (5.03)^8

Let y =(x)^8

and y+∆y=(5.03)^8

Hence Let x =5 and x+∆x= 5.03

→∆x=0.03

Since y=y =(x)^8

→dy/dx=8(x)^7

∆y =(dy/dx) ×∆x =8(x)^7∆x

y+∆y=(5.03)^8 =y+∆x=y+8(x)^7∆x

→ (5.03)^8 =(5)^8++8(5)^7×0.03

=(5)^7{5+0.24) =5.24×(5)^7

Teach Me Derivatives

Assume x in a way such that it is a perfect cube

Let y = ∛x , x = 64 , ∆x = 6

=>dy/dx = 1/(3x^2/3)

=>dy/dx = 1/(3(64)^2/3)

=>dy/dx = 1/48

∆y = ∆x*dy/dx = 6*1/48 = 1/8 = 0.125

∛70 = y + ∆y = ∛64 + 0.125 = 4 + 0.125 = 4.125

Let y = x^8, x = 5, ∆x = 0.03

=>dy/dx = 8x^7

=>dy/dx = 8(5)^7

=>dy/dx = 8*125*125*5 = 1000*625 = 625000

∆y = ∆x*dy/dx = 0.03*625000 = 18750

(5.03)^8 = y + ∆y = 5^8 + 18750 = 125*125*25 + 18750

= 390625+18750 = 409375