Question

Compute the higher order derivative :
f(t) =tcost, find the f ' ' ' ( 0 )

Answer 1

Anyone please solve and explain .

Answer 2

we need to find laplace derivatives?

Answer 3

Then ?

Answer 4

What do you mean Laplace derivative . ?
heres another example
\[g(x) = \sqrt{5-2x} , (Find ) g ' ' ' (2)\]

Answer 5

\[L(\cos(t))=S/S^2+1\]? in this form?

Answer 6

theren is no need of laplace here @aryandecoolest
@EinsteinMorse just use product rule
take
u=t
v=cost

Answer 7

ohk!! @sidsiddhartha
\[dy/dx= u(dv/dx)+v(du/dx)\] ------> use this rule

Answer 8

yeah that'll be enough :)

Answer 9

But its say that Compute the Higher Order Derivative of That ?

Answer 10

yeah just keep on differentiating 3 times as it says
f'''(x)
after that put x=0
and u'll get f'''(0)

Answer 11

yeah at end after differentiating 3 times, put x=0

Answer 12

Then What is the meaning of f ( 0 ) ?

Answer 13

Then in the answer is -3 ?

Answer 14

substitute x=0

Answer 15

I got the derivative 3 times then I substitute the 0 to t , Then I got -3

Answer 16

good. is that correct?. If your procedure is right it should be.

Answer 17

What about the Laplace Derivative . ? im curious about it :)

Answer 18

you know what laplace derivatives are? @EinsteinMorse

Answer 19

I dont know The Laplace derivative and also The partial derivative ?

Answer 20

In this differentiation and integration being performed in the "t"-space are transformed these expressions into the "s"-space expressions...

Answer 21

so if you wanted to calculate tcost from that then
\[L(tcost)=(1/2(1/i-s^2) +1/(i+s^2))\]
\[s^2-1/(s^2+1)^2\]

Answer 22

Thats the Laplace

Answer 23

@aryandecoolest there's a bit mistake in the laplace it will be
\[\frac{ 1-s^2 }{ (1+s^2)^2 }\]

Answer 24

is it @hysenberg ? ) anyways thanks for the verification

Answer 25

What is the Partial Derivative ?

Answer 26

\[d^2f/dx^2= f xx\]
\[d^2f/dxdy=fxy\]
\[d^3x/d^2x dy= fx xy\]

Answer 27

Hows that formula Works :)