How do you differentiate #y=(1+sinx)/(x+cosx)#?

Answer 1

Use the Quotient Rule

#f'(x) = {(x)cos(x)}/(x + cos(x))^2#

The quotient rule:

Given: #f(x) = g(x)/(h(x))#
#f'(x) = {g'(x)h(x) - g(x)h'(x)}/(h(x))^2#
#g(x) = 1 + sin(x)# #g'(x) = cos(x)# #h(x) = x + cos(x)# #h'(x) = 1 - sin(x)#
#f'(x) = {(cos(x))(x + cos(x)) - (1 + sin(x))(1 - sin(x))}/(x + cos(x))^2#
#f'(x) = {(x)cos(x) + cos^2(x) - (1 - sin^2(x))}/(x + cos(x))^2#
#f'(x) = {(x)cos(x) + cos^2(x) - (cos^2(x))}/(x + cos(x))^2#
#f'(x) = {(x)cos(x)}/(x + cos(x))^2#
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Answer 2

To differentiate ( y = \frac{1 + \sin(x)}{x + \cos(x)} ), you can use the quotient rule. The quotient rule states that if you have a function ( u(x) ) divided by another function ( v(x) ), then the derivative is given by:

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

Applying this rule to the given function, we differentiate the numerator and the denominator separately:

[ u(x) = 1 + \sin(x), \quad u'(x) = \cos(x) ] [ v(x) = x + \cos(x), \quad v'(x) = 1 - \sin(x) ]

Now, substituting these into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{1 + \sin(x)}{x + \cos(x)} \right) = \frac{(1 - \sin(x))(x + \cos(x)) - (\cos(x))(1 + \sin(x))}{(x + \cos(x))^2} ]

This expression represents the derivative of the given function.

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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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