# How do I find the derivative of # f(x) = x tan^-1 - ln sqrt(1+x^2)#?

Using the usual rules of diffn., we get,

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To find the derivative of ( f(x) = x \tan^{-1}(x) - \ln(\sqrt{1+x^2}) ), you would apply the chain rule and the derivative rules for trigonometric and logarithmic functions:

[ f'(x) = \frac{d}{dx}[x \tan^{-1}(x)] - \frac{d}{dx}[\ln(\sqrt{1+x^2})] ]

Using the chain rule and the derivative of the natural logarithm function, we get:

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

Simplify and apply the derivative of ( \tan^{-1}(x) ) and ( (1+x^2)^{1/2} ):

[ f'(x) = \tan^{-1}(x) + \frac{x}{1+x^2} - \frac{x}{\sqrt{1+x^2} \cdot \sqrt{1+x^2}} ]

[ f'(x) = \tan^{-1}(x) + \frac{x}{1+x^2} - \frac{x}{1+x^2} ]

[ f'(x) = \tan^{-1}(x) ]

So, the derivative of ( f(x) = x \tan^{-1}(x) - \ln(\sqrt{1+x^2}) ) is ( f'(x) = \tan^{-1}(x) ).

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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.

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