How do you differentiate #sqrtx(sinx+cosx)#?
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To differentiate (\sqrt{x}(\sin{x} + \cos{x})), you can use the product rule of differentiation, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
Let (u = \sqrt{x}) and (v = \sin{x} + \cos{x}). Then, differentiate each function:
(u' = \frac{1}{2\sqrt{x}})
(v' = \cos{x} - \sin{x})
Now, apply the product rule:
((uv)' = u'v + uv')
Substitute the derivatives and the original functions:
((\sqrt{x}(\sin{x} + \cos{x}))' = \frac{1}{2\sqrt{x}}(\sin{x} + \cos{x}) + \sqrt{x}(\cos{x} - \sin{x}))
This expression can be simplified further if needed, but this is the derivative of (\sqrt{x}(\sin{x} + \cos{x})) with respect to (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.
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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