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How do you find the derivative of #y = 6 cos(x^3 + 3)# using the chain rule?

Answer 1

Ezra Adams

#(dy)/(dx) = -18x^2sin(x^3+3)#

#"Let " u=x^3+3 -> (du)/(dx)=3x^2#

#"Let " v= cos(u) -> (dv)/(du) = -sin(u)#

#"Let " y= 6v -> (dy)/(dv) = 6#

Target is #(dy)/(dx)#

By cancelling out #(dy)/(dx) = (dy)/(dv) times (dv)/(du) times (du)/(dx)#

#(dy)/(dx) = (6) times {-sin(u)} times (3x^2)#

#(dy)/(dx) = (6) times (-1) times (3) times {sin(u)} times {x^2}#

#(dy)/(dx) = -18x^2sin(x^3+3)#

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Answer 2

Christopher Allers

To find the derivative of ( y = 6 \cos(x^3 + 3) ) using the chain rule, follow these steps:

Identify the outer function, which is ( \cos(u) ), where ( u = x^3 + 3 ).
Differentiate the outer function with respect to its inner function ( u ), which gives ( -\sin(u) ).
Multiply the result by the derivative of the inner function ( u ) with respect to ( x ), which is ( \frac{d}{dx}(x^3 + 3) = 3x^2 ).

So, applying the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = -6\sin(x^3 + 3) \cdot 3x^2 = -18x^2 \sin(x^3 + 3) ]

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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.