X = acost y = bsint find the second derivative dy / DX of the function

X = acost y = bsint find the second derivative dy / DX of the function

I'm in a hurry. You'd better check it

How do you know whether the left and right derivatives of X are greater than or less than zero?

Are you talking about the left and right derivatives of X, or the left and right derivatives of X?
In the latter case, the basic judgment method is as follows:
1. Is to bring x + Deltax and x-deltax into the original derivative, and then simplify and look at the sign. Deltax > 0
2. A more convenient way is to find the first derivative again and see whether the double derivative value at x is greater than or less than zero. In this way, we can judge whether the first derivative is near a zero field where x is minimal, whether the first derivative is increasing or decreasing, and then we can know the value near it according to the value of the first derivative at X
3. More lazy way, I usually do this, bring in a poor number that is not very far away, and calculate it
4. You can also assist in drawing, which is convenient for judgment

Given the function f (x) = 2x + 3x-12x + 1, find the monotone interval and extreme value of the function

F (x) '= 6 (x ^ 2 + X-2) = 6 (x + 2) (x-1), monotonically increasing interval (- infinity, - 2), (1, + infinity) monotonically decreasing interval [- 2,1], minimum = f (1) = - 6, maximum = f (- 2) = 21

For the image passing point P (0,1) of the even function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e, the tangent equation at x = 1 is y = X-2, and the analytical formula of F (x) is obtained

F (- x) = a (- x) ^ 4 + B (- x) ^ 3 + C (- x) ^ 2 + D (- x) + e = ax ^ 4-bx ^ 3 + CX ^ 2-dx + ef (x) is an even function, so f (x) = f (- x) ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = ax ^ 4-bx ^ 3 + CX ^ 2-dx + e2bx ^ 3 + 2DX = 0. No matter what value x takes, it holds, so B = 0, d = 0, so f (x) = ax ^ 4 + CX ^ 2 + e. the image passes P (0,1), so 1

The tangent equation of the image with odd function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e at x = 1 is y = X-2 Then the analytical formula of F (x) is?

For all x, f (x) = - f (- x), the odd function has ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = - ax ^ 4-x ^ 3-cx ^ 2 + dx-e. ax ^ 4 + CX ^ 2 + e = 0 is constant, so a = C = e = 0f (x) = BX ³+ DX, you can find the tangent point at x = 1, and the tangent point is too (1, - 1), so f (1) = B + D = - 1, you can find f '(1) = 3B + D =

The image of the even function f (x) = ax4 + B3 + CX2 + DX + e passes through (0,1) and the tangent equation y = X-2 at x = 1 to find the analytical formula of F (x)

F (x) image passing (0,1)
f(x)=ax^4+bx^3+cx2+dx+1
F (x) is an even function
f(x)=f(-x)
ax^4+bx3+cx2+dx+1=ax^4-bx^3+cx^2-dx+1
2bx3+2dx=0
b=0 d=0
f(x)=ax^4+cx2+1
f`(x)=4ax^3+2cx
f`(1)=4a+2c=1
f(1)=a+c+1
y=x-2 a+c+1=1-2=-1
a+c=-2
a=5/2 c=-9/2
f(x)=5x^4/2-9x^2/2+1