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What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
What is the derivative of lnlnx?
1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]
What has the exact same variable raised to the same exponent?
An expression that has the same variable raised to the same exponent is x^x. This expression does not have a formal name, however it is worth noting that x^x = e^xlnx.
What is the derivative of cos x raised to the x?
cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(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)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)
What is the antiderivative of 1 over xlnx?
It is ln(ln(x))
What is the derivative of y equals xln x?
(xlnx)' = lnx + 1
What is the ticker symbol for Xilinx?
The ticker symbol for Xilinx is XLNX and it is traded on the Nasdaq.
Is delta v mix 0 for ideal solution?
Yes, it is. delta mixG=nRT(xlnx+...) and delta mixV=dG/dp => delta V=0
What is the derivative of lnlnx?
1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]
What is the Derivative of x to the power of x?
The solution to this is: (xx)'= (elnx to the power of x)'= (exlnx)'= (xlnx)'*exlnx= [x(1/x) + 1(lnx)]*exlnx = (lnx+1)*exlnx= (lnx+1)*xx
What has the exact same variable raised to the same exponent?
An expression that has the same variable raised to the same exponent is x^x. This expression does not have a formal name, however it is worth noting that x^x = e^xlnx.
X the power of x Derivative?
I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx) Here, ^ is power; * = times; ln = natural logratithm ( base e)
What is the derivative of xlnx?
ln(x)+1. This can be found using the product rule: if f(x)=uv, f'(x)=u'v+v'u, so if u is x and v is ln(x), the product rule gives (1)(ln(x))+(x)(1/x)=ln(x)+1.
What is the derivative of y equals xlnx?
Use the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln xUse the product rule.y = x lnxy' = x (ln x)' + x' (ln x) = x (1/x) + 1 ln x = 1 + ln x
What is the derivative of cos x raised to the x?
cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(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)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)
