Question

find ∆y and dy for the function y=x²+ 3x+6,x=10 ∆x=0.01​

Answer 1

Answer:

∆y = 0.2301

dy = 0.23

Step-by-step explanation:

y = x^2 + 3x + 6

y + ∆y = (x + ∆x)^2 + 3(x + ∆x) + 6

x^2 + 3x + 6 + ∆y = x^2 + (∆x)^2 + 2x∆x + 3x + 3∆x + 6

∆y = 2x∆x + 3∆x + (∆x)^2

dy = 2xdx + 3dx

Notice that if ∆x is very small, (∆x)^2 will be negligible and will be approximately zero. In that case, ∆y and dy will be almost equal. The smaller ∆x is, the closer ∆y gets to dy. And that is basically the whole concept of differential calculus.

Answer 2

Answer Given f(x) = x ^ 2 + 3x + 6



x = 10 and Ax=0.01 Delta*y = f(x + Delta*x) - f(x)

= lfloor (x + Delta*x) ^ 2 + 3(x + Delta*x) + 6 ]-(x^ 2 +3x+6) boxed :(a+b)^ 2 =a^ 2 +2ab+b^ 2

= 2x*Delta*x + (Delta*x) ^ 2 + 3Delta*x = (2x + Delta*x + 3) * Delta*x = (23.01)(0.01) = 0.2301

dy= f' * (x) . Delta x

= (2x + 3) * Delta*x

d dx (x^ n )= n * x ^ (n - 1) || d dx (k)=0|

= [2(10) + 3](0.01) = (23)(0.01) = 0.23