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

Answer 1

#(-2/5)(2x-1)^(-6/5)#

Rewrite the problem as:

#(2x-1)^(-1/5)#

Now treat the 2x-1 as an u. The #-1/5# goes down and the u becomes to the power of #-6/5#. So without chain rule it becomes like this:

#(-1/5)(u)^(-6/5)#

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:

#(-2/5)(2x-1)^(-6/5)#

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Answer 2

Cameron Alexander

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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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.