How do I find the derivative of #y = arc cot(x) + (cot x)^-1#?

Answer 1

#y^' = sec^2x - 1/(1 + x^2)#

I'll assume that you don't know what the derivative of #"arccot"(x)# is, so that we can use implicit differentiation to find #(dy)/dx#.

First, start by rewriting your function as

#y = "arccot"(x) + 1/cotx" "#, which is equivalent to
#y = "arccot"(x) + tan(x)#
Rerrange to get #arccos(x)# alone on the right-hand side of the equation
#y - tan(x) = "arccot"(x)#

This is equivalent to

#cot(y -tan(x)) = x " "color(blue)((1))#
Differentiate both sides with respect to #x# to get
#d/dxcot(y - tan(x)) = d/dx(x)#
#-csc^2(y - tan(x)) * [(dy)/dx - sec^2(x)] = 1#
#-csc^2(y - tan(x)) * (dy)/dx + csc^2(y - tan(x)) * sec^2(x) = 1#
Rearrange to get #(dy)/dx# alone on the left-hand side of the equation
#-csc^2(y - tan(x)) * (dy)/dx = 1 - csc^2(y - tan(x)) * sec^2(x)#
#(dy)/dx = (1 - csc^2(y - tan(x)) * sec^2(x))/(-csc^2(y - tan(x)))#
#(dy)/dx = -1/csc^2(y - tan(x)) + (color(red)(cancel(color(black)(-csc^2(y - tan(x))))) * sec^2(x))/color(red)(cancel(color(black)(-csc^2(y - tan(x))))#
#(dy)/dx = sec^2x - 1/csc^2(y - tan(x))#

Use the trigonometric identity

#color(blue)(csc^2x = 1 + cot^2x)#

to write

#(dy)/dx = sec^2x - 1/(1 + cot^2(y - tan(x)))#
You know from equation #color(blue)((1))# that
#cot(y - tan(x)) = x#

which means that the derivative will be

#(dy)/dx = color(green)(sec^2x - 1/(1 + x^2))#
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Answer 2

To find the derivative of ( y = \text{arccot}(x) + (\cot x)^{-1} ), you differentiate each term separately using the chain rule and the power rule for derivatives:

[ \frac{d}{dx} \left( \text{arccot}(x) + (\cot x)^{-1} \right) = \frac{d}{dx} \left( \text{arccot}(x) \right) + \frac{d}{dx} \left( (\cot x)^{-1} \right) ]

[ = -\frac{1}{1 + x^2} + \frac{d}{dx} \left( \frac{1}{\cot x} \right) ]

[ = -\frac{1}{1 + x^2} - \frac{d}{dx} \left( \tan x \right) ]

[ = -\frac{1}{1 + x^2} - \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) ]

[ = -\frac{1}{1 + x^2} - \frac{\cos^2 x + \sin^2 x}{\cos^2 x} ]

[ = -\frac{1}{1 + x^2} - \frac{1}{\cos^2 x} ]

[ = -\frac{1}{1 + x^2} - \sec^2 x ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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