How do you Find the derivative of #y=arctan((1-x)/(1+x) )#?

Answer 1
Let us find the derivative of #{1-x}/{1+x}#.
By Quotient Rule, #({1-x}/{1+x})'={-1cdot(1+x)-(1-x)cdot1}/{(1+x)^2}={-2}/(1+x)^2#
By Chain Rule, #y'=1/{1+({1-x}/{1+x})^2}cdot{-2}/(1+x)^2#
by multiplying the quotients together, #={-2}/{(1+x)^2+(1-x)^2}#
by simplifying the denominator #={-2}/{2(1+x^2)}#
by cancelling out 2's, #=-{1}/{1+x^2}#
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Answer 2

To find the derivative of ( y = \arctan\left(\frac{1-x}{1+x}\right) ), we use the chain rule:

( \frac{dy}{dx} = \frac{d}{dx} \left(\arctan\left(\frac{1-x}{1+x}\right)\right) )

Let ( u = \frac{1-x}{1+x} ). Then, ( \frac{du}{dx} = \frac{d}{dx}\left(\frac{1-x}{1+x}\right) ).

Using the quotient rule, we get:

( \frac{du}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} )

( \frac{du}{dx} = \frac{-1 - x - 1 + x}{(1+x)^2} )

( \frac{du}{dx} = \frac{-2}{(1+x)^2} )

Now, we use the derivative of the arctan function:

( \frac{d}{du}\left(\arctan(u)\right) = \frac{1}{1+u^2} )

Now applying the chain rule:

( \frac{dy}{dx} = \frac{1}{1+\left(\frac{1-x}{1+x}\right)^2} \cdot \frac{-2}{(1+x)^2} )

( \frac{dy}{dx} = \frac{-2}{(1+x)^2 + (1-x)^2} )

( \frac{dy}{dx} = \frac{-2}{1 + 2x + x^2 + 1 - 2x + x^2} )

( \frac{dy}{dx} = \frac{-2}{2(1+x^2)} )

( \frac{dy}{dx} = \frac{-1}{1+x^2} )

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Answer 3

To find the derivative of y = arctan((1 - x)/(1 + x)), we use the chain rule of differentiation. Let u = (1 - x)/(1 + x). Then, y = arctan(u). The derivative dy/dx can be found using the chain rule, which states that if y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx).

First, we find the derivative of y = arctan(u) with respect to u, which is 1 / (1 + u^2).

Next, we find the derivative of u = (1 - x)/(1 + x) with respect to x, which requires the quotient rule. The derivative is ((1 + x)(-1) - (1 - x)(1)) / (1 + x)^2, which simplifies to -2 / (1 + x)^2.

Now, we multiply the derivatives together to get dy/dx: (1 / (1 + u^2)) * (-2 / (1 + x)^2).

Finally, we substitute u = (1 - x)/(1 + x) back into the expression: dy/dx = (-2) / ((1 + (1 - x)^2 / (1 + x)^2) * (1 + x)^2).

This expression can be further simplified if needed, but this is the derivative of y with respect to 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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