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How do you use the chain rule to differentiate #y=-5/(3x^2-4)^6#?

Answer 1

Christopher Agresta

#dy/dx = (180x)/(3x^2-4)^5#

Name #v = x^2# and #u = (3x^2-4)# so that:

#y = -5u^(-6)#

#u = 3v-4#

Applying the chain rule:

#dy/dx = dy/(du) xx (du)/(dv) xx (dv)/dx#

#dy/dx = 30u^(-5) xx 3 xx 2x#

#dy/dx = (180x)/(3x^2-4)^5#

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Answer 2

Cameron Alexander

To differentiate ( y = -5/(3x^2 - 4)^6 ) using the chain rule:

Identify the outer function: ( u = -5u^6 )
Find the inner function: ( u = 3x^2 - 4 )
Differentiate the inner function: ( \frac{du}{dx} = 6(3x^2 - 4)^5(6x) )
Substitute the inner function and its derivative into the outer function: ( \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du} )
Simplify the expression: ( \frac{dy}{dx} = 6(3x^2 - 4)^5(6x) \cdot -\frac{5}{(3x^2 - 4)^6} )
Further simplify if needed.

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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.