The approximate value is calculated by differential calculus 1) arctan 1.02

The approximate value is calculated by differential calculus 1) arctan 1.02

Arctan1.02 = arctan (1 + 0.02) ~ arctan1 + (arctan'1) * 0.02 = P / 4 + [1 / (1 + 1 ^ 2)] * 0.02 = P / 4 + 0.01

How to find the approximate value of √ 63 by differential formula?

Let f (x) = √ X,
Therefore: √ 63 = 8-1 / 16 = 7.9375

In order to calculate the approximate value of 5 √ 270, the differential approximation formula (1 + x) a ≈ 1 + ax can be selected, in which X=

270 to the fifth
270^1/5=[3^5(1+27/3^5)]^1/5
=(3^5)^1/5(1+27/243)^1/5
≈3(1+1/5*27/243)
≈3.07
x=27/243

Inverse function of y = - √ 1-x (x

y=-√(1-x) (x≤1,y≤0)
1-x=y²
x=1-y²
Inverse function: y = 1-x 2 (x ≤ 0)

What is the inverse function of (x-1) of y = 1 + 2Sin (x + 1) Thank you very much for your help

x=1+2sin[(y-1)/(y+1)]
1-2/(y+1)=arcsin[(x-1)/2]
2/(y+1)=1-arcsin[(x-1)/2]
y=2/{1-arcsin[(x-1)/2]}-1

Find the inverse function of y = 1 + 2Sin [(x-1) / (x + 1)]

y=2/{1-arcsin[(x-1)/2]}-1

Inverse function of y = 1 + 2Sin (1-x / 1 + x)

2sin(1-x/1+x)=y-11-x/1+x=arcsin[(y-1)/2]1-x=arcsin[(y-1)/2]+x*arcsin[(y-1)/2]{1+arcsin[(y-1)/2]}x=1-arcsin[(y-1)/2]x={1-arcsin[(y-1)/2]}/{1+arcsin[(y-1)/2]} -1

Finding the inverse function of y = 3 + 2Sin (x-1 / x + 1)

y=arcsin(x-1/5-x)

Y = - the inverse function of 2 parts of - (x-1) (x ∈ R, and X ≠ 1)

Y(X-1)=-2
YX-Y=-2
YX=Y-2
X=(Y-2)/Y
Inverse function: y = (X-2) / X (x ∈ R, and X ≠ 0)

Find the inverse function y = X-2 parts x + 2

y=(x+2)/(x-2)=1+[4/(x-2)]
x=4/(y-1)+2
The inverse function is y = 2 (x + 1) / (x-1), X ≠ 1