How do you differentiate #y = (x+1) (sqrt (2x-1))#?
The simplified condensed solution is:
Now by the product rule we have:
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To differentiate (y = (x+1) \sqrt{2x-1}), use the product rule. The derivative is (\frac{dy}{dx} = (x+1) \frac{d}{dx}(\sqrt{2x-1}) + \sqrt{2x-1} \frac{d}{dx}(x+1)). Differentiate each part separately to get (\frac{dy}{dx} = \frac{1}{2\sqrt{2x-1}}(x+1) + \sqrt{2x-1}). Simplify to get (\frac{dy}{dx} = \frac{x+1}{2\sqrt{2x-1}} + \sqrt{2x-1}).
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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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