I don’t work with arccotangents a lot, but I’ll give it a shot.
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)
y’ = -1/(1+x^2)
And you’re done!
Derivative Of Arccot
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What is the proof that the derivative of arccot (x) = -1/ (1+x^2)?
Let f(x)= cot (x) and its inverse f-inv(x)= arccot (x)
A calculus or analysis text should give you proof of the formula for finding derivative of the inverse, namely:
f-inv’ (x) = 1/(f'(f-inv(x)))
We know the derivative of f(x)= cot(x) is f'(x)= -(csc(x)^2)
This can easily be verified using the fact that
cot(x)= cos (x)/sin(x)
and differentiating using quotient rule and trig idenities.
Now, plug in f'(x) into our formula for f-inv'(x) to see
f-inv'(x) = 1/(-(csc(arccot(x))^2)
Simple algebraic and trigonometric manipulations yeild the required result.
A: 4(x-1)^3 (x+2)^3 + 3(x+2)^4 (x-1)^2 , which simplifies to: 4x^6 + 12x^5 – 12x^4 – 44x^3 +24x^2 + 48x – 32 B: 5 (x^2 + 1)^4 (2x) which equals 10x (x^2 +1)^4 C: 4 ((x^3 – 3x)^3) (3x^2 – 3) which equals 12 (x^2-1) ((x^3 – 3x)^3) D: ( (2x+2) / ((x^2 + 1)^3) ) – ( (6x^3 – 12x^2 – 6x) / ((x^2 + 1)^4) ) wow that was ugly. hope this helps.
The proof is in your brain… try to derive it.