Differential approximation calculation problem When | x | is very small, the approximate formula SiNx ≈ x is derived Let f (x) = SiNx. Take x0 = 0, then SiNx = f (x) ≈ f (x0) + F '(x0) (x-x0) = f (0) + F' (0) X. by F (0) = 0, f '(0) = cos0 = 1, so when | x | is very small, there is SiNx ≈ X Why do we take x0 = 0? Is it necessary to calculate the approximate differential value of | x | → 0? Why?

Differential approximation calculation problem When | x | is very small, the approximate formula SiNx ≈ x is derived Let f (x) = SiNx. Take x0 = 0, then SiNx = f (x) ≈ f (x0) + F '(x0) (x-x0) = f (0) + F' (0) X. by F (0) = 0, f '(0) = cos0 = 1, so when | x | is very small, there is SiNx ≈ X Why do we take x0 = 0? Is it necessary to calculate the approximate differential value of | x | → 0? Why?

Because when | x | is very small, | x | → 0, so take x0 = 0, which is required by this question (determined by the condition when | x | is very small), it is not necessary

Using differential approximation to calculate 990 ^ (1 / 3)

f(x)=x^(1/3),
f(x)≈f(x0)+f`(x0)f(x-x0)
have to
990^(1/3)
=(1000-10)^(1/3)
=10x(1-10/1000)^(1/3)
≈10x[1-1/3x（10/1000）]
≈9.96667

Differential method of composite function

y=f(g(x))
dy/dx=df(g(x))/d(g(x)) * d(g(x))/dx
For example:
y=cos(x^2)
dy/dx=d(cos(x^2))/d(x^2) * d(x^2)/dx
dy/dx=-sin(x^2) * 2x
The differential is: dy = - 2xsin (x ^ 2) DX

Differential method of multivariate composite function Z = XYF (x / y, Y / x) find ∂ Z / ∂ X ∂z/∂x=（∂/∂x)[xyf(x/y,y/x)] =YF (x / y, Y / x) + XY (∂ / ∂ x) f (x / y, Y / x) how did you get this one? =What about YF (x / y, Y / x) + XY [f ① (x / y, Y / x) (1 / y) + F ② (x / y, Y / x) (- Y / x ^ 2)]? =yf(x/y,y/x)+xf①(x/y,y/x)-(y^2/x)f②(x/y,y/x)

（∂/∂x)[xyf(x/y,y/x)]
Take x as a variable and Y is a constant to obtain the derivative of [XYF (x / y, Y / x)]
I can only describe it now,
If you read more examples in the book, you will understand

How to find the differential of compound function? Find the detailed derivation formula

If you're not used to it, you can take the derivative first
Let y = f (U), u = g (V) v = H (x), then y = f (g (H (x)))
y'=f'(u)g'(v)h'(x)
=f'(g(h(x)))g'(h(x))h'(x)
So: dy = f '(g (H (x))) g' (H (x)) H '(x) DX

Differential proof of multivariate compound function If the function u = f (x, y, z) satisfies the identity f (TX, ty, TZ) = T ^ k, f (x, y, z) (T > 0), then f (x, y, z) is called a homogeneous function of degree K. This paper tries to prove the following Euler's theorem on homogeneous functions: a differentiable function f (x, y, z) is a homogeneous function of degree K if and only if: XF_ x (x,y,z)+yF_ y (x,y,z)+zF_ z (x,y,z)=kF(x,y,z)

It is necessary to prove that f (TX, ty, TZ) = T ^ k, f (x, y, z) is always true, and XF is obtained by deriving t from both ends of the equation_ x (tx,ty,tz) + yF_ y (tx,ty,tz) + zF_ Let Z (TX, ty, TZ) = KT ^ (k-1) f (x, y, z), let t = 1, then we can get the conclusion XF_ x (x,y,z) + yF_ y (x,y,z) + zF_ z (...

Approximate value by differentiation Using differential to find the approximate value to the power of 1000 to the tenth power

x^n-(x-Δx)^n≈d(x^n)=nx^(n-1)dx≈nx^(n-1)Δx
So (x - Δ x) ^ n ≈ x ^ n-nx ^ (n-1) Δ x
1000 ^ 0.1 = (1024-24) ^ 0.1 ≈ 1024 ^ 0.1-0.1 * 1024 ^ (- 0.9) * 24 = 2-2.4 / 512 = 1.9953 (Note: 2 ^ 10 = 1024)

Approximate value by differentiation ln1.02

Approximate calculation with derivative
Approximate formula: F (x) = f (x0) + F '(x0) (x-x0)
x=1.02,x0=1,f(x)=lnx
ln1.02=0+1*0.02=0.02

0

f(x)=f(a)+f'(a)(x-a)
1. F (x) = x ^ (1 / 3)
f'(x)=(1/3)x^(-2/3)
x=996,a=1000
f(996)=f(1000)+f'(1000)(996-1000)=1000^(1/3)+(1/3)1000^(-2/3)(-4)
=10-1/75=748/75
2.f(x)=x^(1/6)
f'(x)=(1/6)x^(-5/6)
x=65,a=64
f(65)=f(64)+f'(64)(65-64)=64^(1/6)+(1/6)64^(-5/6)=2-1/192=383/192

The order of differential Engineering (y '') 3 + y (4) cosx = y2inx is

Fourth order