No, they are not brothers.
DX
i think it means 'brothers', which is '弟兄'
dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx dx d dx dx dx dx dx dx dx dx
evolition dx brothers of destruction new age outlaws
Sorry, but if the Jonas Brothers are enabling my peers to spell concert with an 's' ... I hope they don't come over here Dx Oh, and a real answer would be... "Er, probably XD"
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
x5lnx?d/dx (uv)=u*dv/dx+v*du/dxd/dx (x5lnx)=x5*[d/dx(lnx)]+lnx*[d/dx(x5)]-The derivative of lnx is:d/dx(lnu)=(1/u)*[d/dx(u)]d/dx(lnx)=(1/x)*[d/dx(x)]d/dx(lnx)=(1/x)*[1]d/dx(lnx)=(1/x)-The derivative of x5 is:d/dx (xn)=nxn-1d/dx (x5)=5x5-1d/dx (x5)=5x4d/dx (x5lnx)=x5*[1/x]+lnx*[5x4]d/dx (x5lnx)=[x5/x]+5x4lnxd/dx (x5lnx)=x4+5x4lnx
Yes DX is very cool if you dont know what DX is, it is a Tag team on WWEofcourse DX is cool
srry guys, i only know a few, but here they are... FROGGY turns you into a frog ICE CREAM SURPRISE surrounds you with snow flakes SANDWICH makes you smaller...(i would recommend it, it is halarious) and STRAWBERRIES gives you some sort of straw berry hat... XD XD XD XD XD XD XD XD XD XD XD XD DX DX DX DX DX DX DX DX DX DX DX DX
Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
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)
3.9625lnx?The first derivative is:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625lnx)=3.9625*d/dx(lnx)-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)d/dx(lnx)=(1/x)*d/dx(x)d/dx(3.9625lnx)=3.9625*[(1/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(3.9625lnx)=3.9625*[(1/x)*1]d/dx(3.9625lnx)=3.9625*(1/x)d/dx(3.9625lnx)=3.9625/xThe second derivative of 3.9625lnx is the derivative of 3.9625/x=3.9625*x-1:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625*x-1)=3.9625*d/dx(x-1)-The derivative of x-1 is:d/dx(xn)=nxn-1d/dx(x-1)=-1*x-1-1d/dx(x-1)=-1*x-2d/dx(x-1)=-1/x2d/dx(3.9625*x-1)=3.9625*(-1/x2)d/dx(3.9625*x-1)=-3.9625/x2