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How do you differentiate #f(x)=sin(4-x^2) # using the chain rule?

Answer 1

Scarlett Allison

#f' (x)=-2x*cos (4-x^2)#

The solution :

The formula for differentiating #sin u# for any differentiable function #u# is

#d/dx(sin u)=cos u d/dx(u)#

The given: #f(x)=sin (4-x^2)#

#f' (x)=d/dx(sin (4-x^2))=cos (4-x^2) d/dx(4-x^2)#

#f' (x)=cos (4-x^2) *(-2x)#

#f' (x)=-2x*cos (4-x^2)#

God bless....I hope the explanation is useful.

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Answer 2

Samantha Adkison

To differentiate ( f(x) = \sin(4 - x^2) ) using the chain rule:

Identify the outer function: ( \sin(u) ), where ( u = 4 - x^2 ).
Differentiate the outer function with respect to its inner function ( u ), which is ( 4 - x^2 ), yielding ( \cos(u) ).
Differentiate the inner function ( u = 4 - x^2 ) with respect to ( x ), giving ( -2x ).
Apply the chain rule by multiplying the derivatives from steps 2 and 3: ( \cos(u) \times (-2x) ).
Substitute ( u = 4 - x^2 ) back into the expression to get the final result: ( -2x\cos(4 - x^2) ).

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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.