0
Still curious? Ask our experts.
Chat with our AI personalities
Rafa
Blake
RossAdd your answer:
Earn +20 pts
What is the derivative of xy-2y equals 1?
by implicit differentiation you have y+x*dy/dx-2dy/dx=0 solving for dy/dx you'll have dy/dx=y/(2-x) and solving for y in the original equation and plugging it back in, you'll get dy/dx=1/(-x^2 +4x-4) which is your final answer
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
What is the derivative of square root of x-5?
Use Cahin Rule dy/dx = dy/du X du/dx Hence y = ( x-5)^(1/2) Let x - 5 = u Hence dy/du = (1/2)u^(-1/2) du/dx = 1 Combining dy/dx = (1/2)(x-5)^(-1/2)) X 1 dy/dx = 1/ [2(x-5)^(1/2)] Done!!!!
How do you find the derivative of 3 times the square root of x?
Suppose the relation is equal to y then y=3x^1/2 dy/dx = 3*1/2*x^-1/2*1 dy/dx = 3/2*1/x^1/2*1 dy/dx = 3/2x^1/2 First we have retained the constant term i.e. 3 and then we have applied the power rule on x i.e. (nxn-1).
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)
