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Why Are There Only Derivatives to the First Order in the Lagrangian?

Well, the usual physics in classical mechanics is formulated in terms of second-order differential equations. If you are familiar with the process of deriving Euler-Lagrange equations from the Lagrangian then it should be natural that the kinetic term must be proportional to \$(partial_t x)^2\$ to reproduce that.If you would considered more general Lagrangians (which you are certainly free to) you would obtain arbitrarily complicated equations of motions but these would not correspond to anything physical. Nevertheless, some of those equations might describe some mathematical objects (because Lagrangian formalism and calculus of variations is not inherent only to physics but also to lots of other mathematical disciplines)

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Is this proof of power rule for derivatives valid

HINT: You can also use that \$\$x^alpha=e^alphaln(x)\$\$ and use the chain rule

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Application, uses and importance of Hydrocarbon derivatives?

Derivatives in many circumstances seek for suggestion from futures and recommendations. funding homes, banks, businesses all use derivatives to migitate probability. For e.g. an airline employer would purchase oil futures contracts as a thanks to preserve hostile to unexpected oil value will improve.

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Partial Derivatives - Implicit Differentiation

When you taking partial derivative with respect to \$x\$, treat \$y\$ as a constant. Now apply chain rule, Hence \$\$z_x = f'(xy) fracpartial (xy)partial x=f'(xy)\$\$I will leave \$z_y\$ as an exercise

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Trigonometric limits without derivatives [duplicate]

We can proceed without using derivatives in a number of ways. We present two ways herein. APPROACH \$1\$:First, we define the sine function asfrom which we see that \$fracsin(x)x\$ can be written as\$\$fracsin(x)x=sum_n=0^infty frac(-1)^nx^2n(2n1)! tag 1\$\$The series on the right-hand side of \$(1)\$ converges uniformly on any closed, bounded interval. Therefore, we have\$\$beginalign lim_xto 0fracsin(x)x&=lim_xto 0sum_n=0^infty frac(-1)^nx^2n(2n1)! &=sum_n=0^infty ,lim_xto 0left(frac(-1)^nx^2n(2n1)!

ight) &=1 endalign\$\$APPROACH \$2\$:Second, we define the sine function as the inverse function defined by the integralfor \$|x|

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Formulae for PDEs : Commuting derivatives and/or integrals

About the last statement:Total derivatives are sum of partial derivatives. So consecutive partial derivative and total derivative means actually sum of several two consecutive partial derivatives. And we know partial derivatives commute. So, total derivative also commute with partial derivative.

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Geometric meaning of parallelepiped formed by consecutive derivatives

Here is what I think: if \$alpha'''\$ is contained in the osculating plane (that is what your condition says), then \$alpha''\$ is changing (at the given instant) in the osculating plane (I mean, it is not departing from that plane transversally) and the osculating plane itself is stationary at that moment of time. Because of that, the torsion is zero

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derivatives and constants

The difference is that there is no constant term in the first expression. \$3\$ there is coefficient, not a constant term!

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Second order derivatives of a function.

I guess what you are looking for is the following:Let \$vecf(x,y) = beginpmatrix u(x,y) v(x,y) w(x,y)endpmatrix\$. Its "Hessian" is simply the "vector-valued matrix"\$\$ H_vecf = beginpmatrix beginpmatrix u_xx v_xx w_xxendpmatrix & beginpmatrix u_xy v_xy w_xyendpmatrix beginpmatrix u_yx v_yx w_yxendpmatrix & beginpmatrix u_yy v_yy w_yyendpmatrix endpmatrix \$\$where \$ u_xy = fracpartial^2partial xpartial y u\$ etc. Note that each entry of the "matrix" is now a vector in its own right.

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Question on finding derivatives where x=0?

find the derivative first dy/dx = 2(7-3x^2)*6x then plug in 0 for x so the answer is 0 For part 2: you might just be taking the derivative wrong dy/dx = 5*x^448*x^3162*x^2216*x81 so when you plug in 0, the answer is 81 When all else fails, type in the equation into your calculator. put it in Y= Then go to CALC (Push 2nd then TRACE) select the 6th option (dy/dx) then choose at what X value you want to evaluate. press enter and it will give you the answer yeah this one is a little tricky. you know to use product rule when there are two thing multiplied together, and you know to use chain rule because you have a function raised to a power so x(x3)^4 is split into two parts: x * (x3)^4 <--- this is the product rule uv' u'v when taking v' (derivative of second term) you must use chain rule on (x3)^4 thats why both are needed

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