Let the parametric function x = ln (1 + T ^ 2), y = t-arctant. Find (d ^ 2Y) / (DX ^ 2) In addition, if we find dy / DX ^ 2, what about D ^ 2Y / DX?

dy/dx=[1-1/(1+t ²)] / [2t/(1+t ²)]= t/2d ² y/dx ²= (1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1+t ²)]= (1+t ²)/ (4T) I hope it can help you. If the problem is solved, please click the "select as satisfactory answer" button below

Let the function y = y (x) be determined by the equation Xe ^ f (y) = ln2009e ^ y, where f has a second derivative and f '≠ 1, then d ^ 2Y / DX ^ 2

xe^f(y)=ln2009e^y
e^f(y)+xe^f(y)*f'(y)*y' = y'
e^f(y)(1+xf'y')=y'
e^f*f'*y' (1+xf'y')+e^f(f'y'+xy'f''y'+xf'y'')=y

Hurry! Find the second derivative d ^ 2Y / DX ^ 2 of function y determined by the following parametric equation Find the second derivative d ^ 2Y / DX ^ 2 of function y determined by the following parametric equation 1.x=2t-t^2, y=3t-t^3 2.x=f'(t), y=tf'(t)-f(t) (f''(t)≠0)

The first question is 3 / (4-4t)
The second question is 1 / F '' (T)

Find the second derivative d ^ 2Y / DX ^ 2 of the following parametric equation x=t^2/2 y=1-t The answer is 1 / T ^ 3 I don't know how to cut it out I got 1 / T ^ 2

When seeking the second derivative of parametric equation, beginners often make mistakes in the second step
(1) Dy / DX = [dy / dt] / [DX / dt] = - 1 / T this step has fewer errors
(2) D ^ 2Y / DX ^ 2 = D (dy / DX) / DX. Most people who make mistakes regard it as D (dy / DX) / dt. That's your answer
= d(dy/dx) / dt * dt /dx
= 1/t^2 * 1/ t
=1/t^3

Find the second derivative of the parametric equation dy / DX, x = acost, y = bsint

dy/dt=bcost
dx/dt=-asint
dy/dx=-(b/a) *cott
d^2y/dx^2=d(dy/dx)/dx={d(dy/dx)/dt}/(dx/dt)=(b/a) *csc^2 t/-asint=-(b/a^2) *csc^3 t

Why can the second derivative of y be written as d ^ 2Y / DX ^ 2? Can't you write d ^ 2Y / D ^ 2x?

The problem can be understood this way
The sign of the first derivative is dy / DX and the derivative function is y. therefore, D / DX in this sign is equivalent to the derivative symbol (in fact, D / DX is used to represent the derivative symbol in many places)
Since D / DX is the derivative sign, the second derivative of y should be (D / DX) (D / DX) y,
So you can see that there are two D's in the numerator and two DX's in the denominator, so the second derivative is: D ² y/dx ²
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Let the parametric function x = ln (1 + T ^ 2), y = t-arctant. Find (d ^ 2Y) / (DX ^ 2) In addition, if we find dy / DX ^ 2, what about D ^ 2Y / DX?