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How do you use the chain rule to differentiate #root9(-cosx)#?

Answer 1

Luke Alvarez

#= (sin(x))/(9(-cos(x))^(8/9))#

We have #y = (-cos(x))^(1/9)#

Let #u = -cos(x) implies (du)/(dx) = sin(x)#

#y = u^(1/9)# so using power rule:

#(dy)/(du) = 1/9(u)^(-8/9)#

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

#(dy)/(dx) = 1/9(-cos(x))^(-8/9)*sin(x)#

#= (sin(x))/(9(-cos(x))^(8/9))#

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Answer 2

Emily Ammerman

To differentiate √9(-cos(x)) using the chain rule, follow these steps:

Let u = -cos(x).
Find the derivative of u with respect to x: du/dx = sin(x).
Let v = √9u.
Find the derivative of v with respect to u: dv/du = 1/2(9u)^(-1/2) = 1/(2√9u) = 1/(6√u).
Apply the chain rule: dv/dx = dv/du * du/dx.
Substitute back u = -cos(x) and simplify the expression to get the final answer.

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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.