Question

2لے
12. Find the approximate value of Jo 037 using differential
approximate?​

Answer 1

Step-by-step explanation:

Approx value of

0.082

Let function y=f(x)=

x

Let x=0.081x+Δx=0.082

Δx=0.001

for x=0.081y=0.009

Let dx=Δx=0.001

y=

x

=

2

x

1

(

dx

dy

)

x=0.081

=

2×0.009

1

=

0.081

1

=55.5

dy=

dx

dy

dx dy=55.5×0.001

Δy=0.0555

Hence

0.082

=0.009+0.0555=0.065

Answer 2

Answer:

Step-by-step explanation:

Approx value of

0.082

Let function y=f(x)=

x

Let x=0.081x+Δx=0.082

Δx=0.001

for x=0.081y=0.009

Let dx=Δx=0.001

y=

x

=

2

x

1

(

dx

dy

)

x=0.081

=

2×0.009

1

=

0.081

1

=55.5

dy=

dx

dy

dx dy=55.5×0.001

Δy=0.0555

Hence

0.082

=0.009+0.0555=0.065