# How do you differentiate #y = (sqrt(cos x))/(5lnx)#?

use quotient rule:

(see this proof or this)

when you simplify you should get something like:

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To differentiate the given function ( y = \frac{\sqrt{\cos x}}{5\ln x} ), you would use the quotient rule of differentiation:

[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ]

Where ( u = \sqrt{\cos x} ) and ( v = 5\ln x ).

Now, differentiate ( u ) and ( v ) separately:

[ u' = \frac{-\sin x}{2\sqrt{\cos x}} ] [ v' = \frac{5}{x} ]

Now apply the quotient rule:

[ \frac{d}{dx}\left(\frac{\sqrt{\cos x}}{5\ln x}\right) = \frac{\frac{-\sin x}{2\sqrt{\cos x}} \cdot (5\ln x) - \sqrt{\cos x} \cdot \frac{5}{x}}{(5\ln x)^2} ]

Simplify the expression:

[ \frac{-5\sin x \ln x}{2x\sqrt{\cos x}} - \frac{5\sqrt{\cos x}}{x(5\ln x)^2} ]

This is the derivative of the given function.

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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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