Answer:
a)
[tex]\triangle y = 1.56[/tex]
b) Dy=1.2
c) See attachment
Step-by-step explanation:
a)Let
[tex] \triangle x[/tex]
be a small change in x, then the corresponding change in y is
[tex] \triangle y= f(x + \triangle x) - f(x)[/tex]
We have
[tex]f(x) = {x}^{2} [/tex]
This implies that:
[tex] \triangle y= (x + \triangle x)^{2} - {x}^{2} [/tex]
We substitute
[tex]x = 1 \: and \: \: \triangle x = 0.6[/tex]
This gives
[tex] \triangle y= (1 + 0.6)^{2} - {1}^{2} \\ \triangle y = {1.6}^{2} - 1 = 1.56[/tex]
b) Let
[tex]y = f(x)[/tex]
[tex] \frac{dy}{dx} = f'(x)[/tex]
This implies that:
[tex]dy = f'(x)dx[/tex]
with dx=0.6 , and x=1,
[tex]dy = 2x dx[/tex]
[tex]dy = 2(1)(0.6) = 1.2[/tex]