What is the derivative of #arctan(x)/(1+x^2)#?
If we differentiate implicitly, we get
Let's now use the quotient rule to differentiate the entire expression.
Hopefully this helps!
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To find the derivative of ( \frac{\arctan(x)}{1+x^2} ), use the quotient rule. Let ( u = \arctan(x) ) and ( v = 1 + x^2 ). Then:
[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} ]
[ \frac{du}{dx} = \frac{1}{1+x^2} ] [ \frac{dv}{dx} = 2x ]
Substitute these values into the quotient rule formula:
[ \frac{d}{dx}\left(\frac{\arctan(x)}{1+x^2}\right) = \frac{(1+x^2)\left(\frac{1}{1+x^2}\right) - \arctan(x)(2x)}{(1+x^2)^2} ]
[ = \frac{1 + x^2 - 2x^2\arctan(x)}{(1+x^2)^2} ]
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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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