Derivative of implicit function ye x lny 1, find dy dx, how to do it

The partial derivative of both sides of the equation with respect to x yields:

y'e^x +ye^x +y'/y=0

i.e. (e x +1 y)y'+ye x=0 to get y'= -y e x (1+ye x) is obtained by ye x+lny=1: ye x =-lny+1 so y' = -y²e^x /(1+ye^x) =-y(-lny+1)/[1-lny+1]=(ylny-y)/(2-lny)

Both sides are derivative of x at the same time.

Use the product rule + composite to find the derivative.

dy/dx)e^x+ye^x+(1/y)*dy/dx=0dy/dx)(e^x+1/y)=-ye^xdy/dx=-ye^x/(e^x+1/y)ye^x=1-lny

e^x=(1-lny)/y

Substitute for the return. dy/dx=-(1-lny)/((1-lny)/y+1/y)y(lny-1)/(2-lny)

y=x+lny

Both sides are guided at the same time.

dy/dx=1+1/y*dy/dx

1-1/y)dy/dx=1

dy/dx=1/(1-1/y)=y/(y-1)Extended MaterialsFor a case that has been determined to exist and is derivable, we can use the chain rule of derivatives of composite functions to find the derivative. Derivatives of x are taken on both sides of the equation, and since y is actually a function of x, you can get a function of x directly'and then simplify to get y'expression.

The following methods can generally be used to solve the derivatives of implicit functions:

Method 1: First, the implicit function is converted into an explicit function, and then the derivative is obtained by using the method of derivative of the explicit function.

Method 2: Derivative of x on the left and right sides of the implicit function (but be careful to treat y as a function of x);

Method 3: Use the invariant properties of the first-order differential form to find the derivatives of x and y respectively, and then move the values to obtain the values.

Both sides are derived from x at the same time, i.e., dy dx=1+(1 y)*dy dx(1-1 y)dy dx=1

dy dx=1 (1-1 y)=y (y-1)Note: The derivative of lny to x is a problem of deriving a composite function, first to y, and then to x, which is (1 y)*dy dx of the above equation

Finding the derivative of y on both sides gives dx dy=1-1 y; So dy dx=y (y-1).

x=yln(xy), the derivatives of x at both ends of the equation, 1=dy dx+y[1 ln(xy)][y+x(dy dx)]=dy dx+y ln(xy)+xdy chaotic dx, sorted out to get a hail.

Dy dx) (1+x)=1 y ln(xy), i.e. dy dx={[ln(xy)-y] [(1+x)ln(xy)].

x=yln(xy), the equation is that the two fierce gods are dressed as derivatives for x, 1=dy dx+y[1 hengzhizao ln(xy)][y+x(dy dx)]=dy dx+y ln(xy)+xdy dx, sorted out.

Dy dx) (1+x)=1 y ln(xy), i.e., dy dx={[ln(xy)-y] [1+x)ln(xy)],1,

Directly find the derivative of x on both sides, get 1 y*(-1 y2)*dy dx=1 xy*(y+xdy dx) Let's get it, 3, shilling k=dx dy, then both sides derive x respectively, 1 dy dx=1 (xy)(y+x*dy dx) and then merge the same terms to get k-1 (ky)=1 x k 2*y-ky x=1 and then reverse the solution k then dy dx=1 k,0,x y=ln(xy) find the derivative of the implicit function y dy Lu Songxi dx

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Hidden letter ode to the side of the number of wild slips to guide and remember.

F(y) is derived from x'(y) *y'

Formula x +lny-y=0

Then the derivative of x is obtained.

2x+y'/y -y'=0

So the ship was set off'= 2xy/(y-1)

Solution: The curve equation is trapped by x lny-y=0, and there is 2x y'/y-y'=0, shirt cautious y'-y'or imitation y=2x,y'(1-1 y) = 2x, get: y'=2xy/(y-1)

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Finding the derivative of the implicit function x+y+2xy=1 dy dx Don't worry, I'm also manual, I'm writing it and sending it to you.

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Find the differential dy, which is the function y=e cosx

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Summary. Dear - Find the derivative of the implicit function x+y+2xy=1 dx dy--- find the derivative of the implicit function determined by the equation exy+x2y=1 [dy dx] swallow171 1 year ago 1 report has received 1 report like carpex seedlings a total of 21 problem rate: 81% Report solution ideas:

Derivative of the two sides of the equation exy+x2y=1 directly for x, and at the same time treat y as a function of x, you can find the derivative of the equation exy+x2y=1 on both sides of x, and get exy(y+xdydx)+2xy+x2dydx 0 [dy dx 2xy+yexyx(1+exy)].

Dear - Finding the Hidden Function Pei Ming x+y+2xy=1 Derivative dx dy--- Finding the Derivative of the Implicit Function [dy dx] by the equation exy+x2y=1 and determining the derivative of the implicit function [dy dx] Matching Empty Announcement swallow171 1 year ago Received 1 report like carpex seedlings with a total of 21 problem rate: 81% Report solution idea: The equation exy+x2y=1 is directly derived from x, and at the same time, the equation exy+x2y=1 is directly derived from x, and at the same time, y is regarded as a function of x, and the derivative of xto get exy(y+xdydx)+2xy+x2dydx 0 [dy dx 2xy+yexyx(1+exy)]

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2xdy+2ydx+dy=0 (2x+1)dy=-2ydx dy dx=-2y (2x+1) (2x+1)y=1 y=1 y=1 (2x+1) dy dx=-2 (2x+1) 2

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y=x+lny

Both sides are guided at the same time.

dy dx=1+1 y*dy Yu Yun dx

1-1/y)dy/dx=1

dy dx=1 quietly(1-1 qingzaoy)=y (y-1).