How do you differentiate #f(x)=cos(xe^(x) ) # using the chain rule?
Using product rule #(xe^x)' =x(e^x)'+e^x(x)' =xe^x+e^x =( 1 + x )e^x#.
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To differentiate ( f(x) = \cos(xe^x) ) using the chain rule, follow these steps:
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Identify the outer function ( u ) and the inner function ( v ). ( u = \cos(v) ) where ( v = xe^x ).
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Differentiate the outer function ( u ) with respect to its variable ( v ). ( \frac{du}{dv} = -\sin(v) ).
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Differentiate the inner function ( v ) with respect to the original variable ( x ). ( \frac{dv}{dx} = e^x + xe^x ).
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Apply the chain rule formula: ( \frac{df}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} ).
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Substitute the derivatives found in steps 2 and 3 into the chain rule formula. ( \frac{df}{dx} = -\sin(xe^x) \cdot (e^x + xe^x) ).
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Simplify the expression if necessary.
So, ( f'(x) = -\sin(xe^x) \cdot (e^x + xe^x) ).
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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.
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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