How do you find the derivative of #(tanx)^-1#?

Answer 1

I would rewrite the function before differentiating

#(tanx)^-1 = 1/tanx = cotx#
Now use the memorized #d/dx(cotx) = -csc^2x# or rewrite another step
#cotx = cosx/sinx# and use the quotient rule to get
#d/dx(cotx) = (-sin^2x-cos^2x)/sin^2x = (-1)/sin^2x = -csc^2x#
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Answer 2

To find the derivative of ((\tan(x))^{-1}), you can use the chain rule. Let (y = (\tan(x))^{-1}). Then, differentiate (y) with respect to (x), and apply the chain rule.

[\frac{dy}{dx} = -\frac{1}{(\tan(x))^2} \cdot \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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