Question

prove that D[arccotx]=-(1/(1+x^2)

Answer 1

whats captial D mean?

Answer 2

Derivative

Answer 3

It's a shorter way of saying dy/dx essentially

Answer 4

ah ok, ty

Answer 5

can you use the definition of a derivative to prove it?

Answer 6

If I knew the properties of arccot, then yeah....but I don't

Answer 7

ah ok

Answer 8

1. Write down y=arccot(x).
2. Draw a picture of a right triangle with base x, height 1, and angle y (between the base and hypotenuse). The hypotenuse will therefore be sqrt(1+x^2). This triangle represents the relationship between x and y and will be important later on.
3. Take the cotangent of both sides to give x=cot(y).
4. Rewrite cot(y) as cos(y)/sin(y).
5. Implicitly differentiate the function. I hope you know how to do this. If you don't, the basic rule is to differentiate both x and y terms as you normally would, except all terms with a y in them are multiplied by y' (where y' is dy/dx). Here's how it's done with implicit differentiation and the quotient rule:
x = cos(y)/sin(y)
1 = (sin(y)*(-sin(y)*y') - cos(y)*cos(y)*y')/sin^2(y)
1 = (-sin^2(y)*y' - cos^2(y)*y')/sin^2(y)
1 = y'*(-sin^2(y) - cos^2(y))/sin^2(y)
1 = -y'/sin^2(y) (This is thanks to the identity sin^2(y)+cos^2(y)=1)
6. Now get the derivative term-- the y' term-- on one side of the equation:
y' = -sin^2(y)
7. Look at your drawing of the triangle. To get a proper value for the sin, you must divide the length of each side by sqrt(1-x^2) so that the hypotenuse is length 1. Now the base of the triangle is x/sqrt(1+x^2) and the height is 1/sqrt(1+x^2).
8. Remember that the sin of a function is the length of a side opposite the angle on a right triangle with hypotenuse 1. In other words, the sin of y is 1/sqrt(1+x^2)
9. Substitute that into your equation
y' = -sin^2(y)
to give
y' = -1/(1+x^2)