What is the derivative of #y=tan(x)/x#?

Answer 1
This function, in the form of #y = f(x) = g(x)/(h(x))#, is a perfect candidate for using the quotient rule.
The quotient rule states that the derivative of #y# with respect to #x# can be solved with the following formula:
Quotient rule: #y'= f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x)^2)#

In this problem, we can assign the following values to the variables in the quotient rule:

#g(x) = tan(x)# #h(x) = x#
#g'(x) = sec^2(x)# #h'(x) = 1#

If we plug these values into the quotient rule, we get the final answer:

#y' = (sec^2(x) * x - tan(x) * 1)/x^2 = (xsec^2(x) - tan(x))/x^2#
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Answer 2

To find the derivative of ( y = \frac{\tan(x)}{x} ), we can use the quotient rule.

Let ( u = \tan(x) ) and ( v = x ).

Now, ( \frac{du}{dx} = \sec^2(x) ) (derivative of ( \tan(x) )) and ( \frac{dv}{dx} = 1 ).

Applying the quotient rule ( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ), we get:

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

( \frac{dy}{dx} = \frac{x\sec^2(x) - \tan(x) - \tan(x)}{x^2} )

( \frac{dy}{dx} = \frac{x\sec^2(x) - 2\tan(x)}{x^2} )

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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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