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How do you use the chain rule to differentiate #y=8(x^4-x+1)^(3/4)#?

Answer 1

Emilia Aguilar

#dy/dx=6(x^4-x+1)^(-1/4) +(4x^3-1)#

Given -

#y=8(x^4-x+1)^(3/4)# #dy/dx=[8*3/4(x^4-x+1)^((3/4)-1)]xxd/dx(x^4-x+1)#

#dy/dx=6(x^4-x+1)^(-1/4)(4x^3-1)#

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Answer 2

Adam Adkins

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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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.