# How do you find the derivative of # f(x)=sin e^(4x) + cos e^(4x)# using the chain rule?

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To find the derivative of ( f(x) = \sin(e^{4x}) + \cos(e^{4x}) ) using the chain rule, first recognize that both terms inside the sine and cosine functions are composite functions. Apply the chain rule separately to each term. The chain rule states that if ( u ) is a differentiable function of ( x ) and ( f ) is a differentiable function of ( u ), then the derivative of ( f(u) ) with respect to ( x ) is ( f'(u) \cdot u' ).

For the first term, ( \sin(e^{4x}) ), let ( u = e^{4x} ). Then ( \frac{du}{dx} = 4e^{4x} ). Applying the chain rule, the derivative of ( \sin(e^{4x}) ) with respect to ( x ) is ( \cos(e^{4x}) \cdot 4e^{4x} ).

Similarly, for the second term, ( \cos(e^{4x}) ), let ( u = e^{4x} ). Then ( \frac{du}{dx} = 4e^{4x} ). Applying the chain rule, the derivative of ( \cos(e^{4x}) ) with respect to ( x ) is ( -\sin(e^{4x}) \cdot 4e^{4x} ).

Therefore, the derivative of ( f(x) = \sin(e^{4x}) + \cos(e^{4x}) ) with respect to ( x ) is:

[ f'(x) = \cos(e^{4x}) \cdot 4e^{4x} - \sin(e^{4x}) \cdot 4e^{4x} ]

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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.

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.

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.

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