for enjoys
Pocket Tanks is a 1-2 player computer game for Windows and Mac OS X, and more recently, the iPhone, created by Blitwise Productions, developer of Super DX-Ball and Neon Wars.
5ex+2?d/dx(u+v)=du/dx+dv/dxd/dx(5ex+2)=d/dx(5ex)+d/dx(2)-The derivative of 5ex is:d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex)=5*d/dx(ex)-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=(5*d/dx(ex))+(0)d/dx(5ex+2)=5*d/dx(ex)-The derivative of ex is:d/dx(eu)=eu*d/dx(u)d/dx(ex)=ex*d/dx(x)d/dx(5ex+2)=5*(ex*d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(5ex+2)=5*(ex*1)d/dx(5ex+2)=5*(ex)d/dx(5ex+2)=5ex5ex+2?d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex+2)=5*d/dx(ex+2)-The derivative of ex+2 is:d/dx(eu)=eu*d/dx(u)d/dx(ex+2)=ex+2*d/dx(x+2)d/dx(5ex+2)=5*(ex+2*d/dx(x+2))-The derivative of x+2 is:d/dx(u+v)=du/dx+dv/dxd/dx(x+2)=d/dx(x)+d/dx(2)d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=5*[ex+2*(1+0)]d/dx(5ex+2)=5*[ex+2*(1)]d/dx(5ex+2)=5*[ex+2]d/dx(5ex+2)=5ex+2
It is already out. The sequel to that game is called Sonic Adventure 2: Battle. Dx is a re release of sonic adventure, and they also re released sonic adventure 2.
The game will automatically save itself.
One way to get a lot of Master Balls is by winning the grand prize at the game's Lottery however another one is by having at least 2 games, one of which being a primary game that you don't plan on restarting and the other being a secondary game that you plan on restarting, the other way is to use the secondary game until you get to the part in which you get the Master Ball and then you attach it to your Pokémon and then you trade that Pokémon to your primary game and then you delete the save file on your secondary game and start a new game so you can get another Master Ball in a new save file.
The definition of the derivative, at a point X = x is the limit, as dx tends to 0, of [f(x+dx)-f(x)]/dx. In this case, therefore, it is lim[3*(x+dx)+2 - (3*x+2)]/dx = lim[3x + 3*dx +2 - 3x - 2]/dx = lim[3*dx/dx] = lim[3] = 3.
√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]
k*tan[√(4x)]=k*tan[(4x)1/2], where k is a constant:Multiplying by a constant, multiply the derivative of u by the constant c: d/dx d/dx(cu)=c*du/dxd/dx(k*tan[(4x)1/2])=k*d/dx(tan[(4x)1/2])-The derivative of tan[(4x)1/2] is:d/dx(tan u)=sec2(u)*d/dx(u)d/dx(tan[(4x)1/2])=sec2([(4x)1/2])*d/dx([(4x)1/2])d/dx(tan[(4x)1/2])=sec2(2√x)*d/dx([(4x)1/2])d/dx(k*tan[(4x)1/2])=k*{sec2(2√x)*d/dx([(4x)1/2])}-The derivative of (4x)1/2 is:Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(4x)1/2=(1/2)*(4x)1/2-1*d/dx(4x)d/dx(4x)1/2=(1/2)*(4x)-1/2*4d/dx(4x)1/2=4/[2(4x)1/2]d/dx(4x)1/2=4/[2(2√x)]d/dx(4x)1/2=4/[4√x]d/dx(4x)1/2=1/(√x)d/dx(k*tan[(4x)1/2])=k*sec2(2√x)*(1/√x)d/dx(k*tan[(4x)1/2])=[k*sec2(2√x)]/√x
25x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(25x)=25x*ln(2)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(25x)=95x*ln(2)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(25x)=25x*ln(2)*(5*1)d/dx(25x)=25x*ln(2)*(5)-25x can simplify to (25)x, which equals 32x.d/dx(95x)=32x*ln(2)*(5)
If you save it properly then yes it will.
d/dx(x + 2) = d/dx(x) + d/dx(2) = 1 + 0 = 1