The power of 2x square of y = e

The power of 2x square of y = e

y=e^(2x^2)
f(x)=e^x,g(x)=2x^2
f(g(x))'=f'(g(x))*g'(x)
=e^(2x^2)*4x

Given that the tangent equation of the curve y = f (x) at x = 1 is 2x-y + 1 = 0, the differential dy of x = 1 is obtained

If the tangent equation at x = 1 is y = 2x + 1, then the tangent slope is 2,
That is, f '(1) = 2
Dy = f '(1) DX = 2DX at x = 1

U = x Λ (y + Z2), find the first order partial derivative and total differential (using the form invariance of total differential)

Find f (x) = x3-3x K, G (x) = (2kx-k) / (X22) 1 / 2 × 2 / 3 × 3 / 4 × 4 / 5 × ×9
X^2-3XY 2Y^2an=1/n * (-1)^[(3 n)/2]=-(-1)^(n/2 1/2) /n

A (5,2) vs B (- 3,0) a = b = vs f (x) f (x 1) = 8x 7ab = AC = 3cos = 1 / 9

Y = the domain of the radical (√ 3 - cos x)
Sorry, it should be tan X

According to the image of TaNx, X ∈ (K π - π 2, K π + π 3] K ∈ Z

Find (1-cos X / 2) under y = root

If √ (1-cos X / 2) is defined, then (1-cos X / 2) ≥ 0 under the root sign, that is to say, cos X / 2 ≤ 1 is required;
So the domain of definition is all real numbers, that is (- ∞, + ∞)

It is known that f (1-3x) domain is (- 1,3) to find f (2x-1) + F (5-2x) domain

In y = f (1-3x), - 1-9
4>1-3x>-8
In F (2x-1) + F (5-2x)
-8

It is known that the domain of function f (x) is [- 1,3]. Find the domain of function f (x) = f (2x + 1) - 4f (3x-2)

A:
The domain of F (x) is [- 1,3]
The definition field of F (x) = f (2x + 1) - 4f (3x-2) satisfies:
-1

If the domain of F (1-2x) is known to be [- 1,5], find the domain of F (3x-5); 4. If the domain of F (1-2x) is known to be [- 1,5], find F（
2. Given that the domain of F (x) is [- 1,5], find the domain of F (1 + m) + F (1-m).

-1﹤1-2x﹤5
-2﹤x﹤1
So - 6 ＜ 3x ＜ 3 - 11 ＜ 3x-5 ＜ - 2
The domain of F (3x-5) is (- 11, - 2)
-1＜1-2x＜5
-2＜x＜1
Because both 1 + m and 1-m are equivalent to X,
So the domain of F (1 + m) + F (1-m) is (- 2,1)

Given that the domain of F (2x-1) is [0,1], find the domain of F (1-3x)
To detailed process, and reason, why is the answer, do not understand ah

The definition field refers to the value of X;
For these two functions, what remains unchanged is the corresponding relation F;
Let t = 2x-1;
X ∈ [0,1), then t ∈ [- 1,1];
The corresponding relation f remains unchanged and the original image of F remains unchanged;
1-3x∈[-1,1);
x∈(0,2/3];