Finding the derivative of y = 1 / (1-x)

Finding the derivative of y = 1 / (1-x)

y=(1-x)^(-1)
So y '= - 1 * (1-x) ^ (- 1-1) * (1-x)'
=-(1-x)^(-2)*(-1)
=1/(1-x)^2

What is the derivative of [(1 + x ^ 2) ^ - 1]?

y=1/(x^2+1)
It can be transformed into:
y(x^2+1)=1
The derivative of both sides is obtained
y'(x^2+1)+y(2x)=0
y'=-2xy/(x^2+1)
=-2x/(x^2+1)^2

What is the derivative of X + 1 / x + 2?

1-x^-2

Why is the derivative of x 1

Because the definition of derivative is LIM (H - > 0) [f (x + H) - f (x)] / h
So f (x) = x, the derivation of X is LIM (H - > 0) [x + H - x] / h = 1

What is the derivative of y = x + 16 △ √ (x-1) (x > 1)

y=x+16(x-1)^(-1/2)
So y '= 1 + 16 * (- 1 / 2) * (x-1) ^ (- 1 / 2-1)
=1-8*(x-1)^(-3/2)
=1-8/[(x-1)√(x-1)]

Derivation of e ^ (XY) + sin (x + y) + 1 = 0 implicit function
Derivation of e ^ (XY) + sin (x + y) + 1 = 0 implicit function

e^(xy)+sin(x+y)+1=0[e^(xy)]'+[sin(x+y)]'+1'=0'e^(xy)*(xy)'+cos(x+y)*(x+y)'+0=0e^(xy)(x'y+xy')+cos(x+y)*(x'+y')=0e^(xy)(y+xy')+cos(x+y)(1+y')=0ye^(xy)+xe^(xy)y'+cos(x+y)+cos(x+y)y'=0[xe^(xy)+cos(x+y)]y...

Derivation of y = ln (XY) + 1 implicit function

y=ln(xy)+1
Both sides of X at the same time to obtain the derivative
y '=1/(xy)·(xy)'
y '=(y+xy ')/(xy)
y '=y '/y+1/x
y '(1-1/y)=1/x
y '=y(y-1)/x

Let y = y (x) be an implicit function determined by sin (XY) = LNX + ey + 1, then y '(0)=______ ．

In the equation, let x = 0, 0 = lney (0) + 1, then we can get y (0) = E2, we can get cos (XY) (y + XY ′) = 1x + e − y ′ y, we can substitute x = 0, y (0) = E2, there is E2 = 1E − y ′ (0) E2, that is, y ′ (0) = e-e4

An example of derivation of implicit function e ^ y + xy-e = 0
It is said in the book that e ^ y * y '+ y + X * y' = 0 is the derivative of X on the left side of the equal sign. How does he make the E on the left side of the equal sign become + X * y 'after the derivation? Doesn't it mean that the constant derivative is equal to zero? Why is it so? After the book, practice the first y ^ 2-2xy + 9 = 0, and there is a constant on the left side of the equal sign. How can we derive x on the left side of the equal sign?

So (e ^ y) '= e ^ y * y' (XY) 'is the same as (XY)' = x '* y + X * y' = y + X * y'e is a constant, and the derivative = 0, so e ^ y + xy-e = 0, and then e ^ y * y '+ y + X * y' = 0y '= - Y / (e ^ y + x) y ^ 2-2xy + 9 = 0

An example of derivation of implicit function e ^ y + xy-e = 0
It is said in the book that e ^ y * y '+ y + X * y' = 0 is the derivative of X on the left side of the equal sign. How does he make the E on the left side of the equal sign change to + X * y 'after derivation? Doesn't it mean that the constant derivative equals zero? Why is this? Who can explain? After the book, there are constants on the left side of the first y ^ 2-2xy + 9 = 0 equal sign. How can we derive x on the left side of the equal sign?

e^y+xy-e=0
E ^ y: e ^ y * y '
XY: y + X * y '
E is the derivative of X: 0
Add the results
e^y*y'+y+x*y'=0
y^2-2xy+9=0
2y*y'-2y-2xy'=0
y'=y/(y-x)