Question

for the function y= x^3,if x=5 and delta x=0.01,find delta y

Answer 1

$y = {x}^{3} = >$
on differentiating on both sides, we get
$\frac{dy}{dx} = 3 {x}^{2}$
=>
$dy = 3 {x}^{2} dx = 3 {(5)}^{2} (0.01) = 0.75.$
hence dy =0.75 .Hope it helps you...Mark me as Brainliest if you think i am worthy of it.

Answer 2

Answer:0.75

Step-by-step explanation:

Given

y=x^3

dy/dx=3x^2

dy=3x^2.dx

=3(5^2)x0.01

=75x0.01

=0.75

dy=0.75

Thank you hope you guys get this.