How do you use the chain rule to differentiate #y=8(x^4x+1)^(3/4)#?
#dy/dx=6(x^4x+1)^(1/4) +(4x^31)#
Given 
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To differentiate ( y = 8(x^4  x + 1)^{\frac{3}{4}} ) using the chain rule:

Identify the outer function and the inner function. Outer function: ( u = 8u^{\frac{3}{4}} ) Inner function: ( v = x^4  x + 1 )

Differentiate the outer function with respect to ( u ). ( \frac{d}{du} (u^{\frac{3}{4}}) = \frac{3}{4}u^{\frac{3}{4}1} = \frac{3}{4}u^{\frac{1}{4}} )

Differentiate the inner function with respect to ( x ). ( \frac{dv}{dx} = 4x^3  1 )

Combine the derivatives using the chain rule: ( \frac{dy}{dx} = \frac{d}{dx}(8v^{\frac{3}{4}}) = \frac{du}{dv} \times \frac{dv}{dx} ) ( = \frac{3}{4}u^{\frac{1}{4}} \times (4x^3  1) )

Substitute ( u ) and ( \frac{du}{dx} ) back in: ( = \frac{3}{4}(8x^4  8x + 8)^{\frac{1}{4}} \times (4x^3  1) )

Simplify the expression if necessary.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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