# How do you use the chain rule to differentiate #y=1/root5(2x-1)#?

Rewrite the problem as:

But then chain rule says you have to take the derivative of the inside which we called u and multiply it to our result. u was 2x-1. The derivative of u is 2. So we simply multiply by 2 and plug 2x-1 back in for u and get:

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To differentiate ( y = \frac{1}{\sqrt{5}} (2x - 1) ) using the chain rule:

- Identify the inner function: ( g(x) = 2x - 1 )
- Identify the outer function: ( f(u) = \frac{1}{\sqrt{5}}u ), where ( u = 2x - 1 )
- Calculate ( f'(u) ) and ( g'(x) )
- ( f'(u) = \frac{d}{du}\left(\frac{1}{\sqrt{5}}u\right) = \frac{1}{\sqrt{5}} )
- ( g'(x) = \frac{d}{dx}(2x - 1) = 2 )

- Apply the chain rule: ( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) )
- Substitute the derivatives and simplify:

[ \frac{dy}{dx} = \frac{1}{\sqrt{5}} \cdot 2 = \frac{2}{\sqrt{5}} ]

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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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