How do I find #f'(x)# for #f(x)=3^-x# ?
First, step is a change of base:
rearrange and you will get the same answer as the first line.
The other option is to use the general exponential differentiation rule (if you can remember it):
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To find ( f'(x) ) for ( f(x) = 3^{-x} ), you can use the chain rule. The chain rule states that if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.
So, for ( f(x) = 3^{-x} ), the outer function is ( g(u) = 3^u ) and the inner function is ( u(x) = -x ).
Now, the derivative of the outer function ( g(u) = 3^u ) with respect to ( u ) is ( g'(u) = 3^u \ln(3) ).
The derivative of the inner function ( u(x) = -x ) with respect to ( x ) is ( u'(x) = -1 ).
Now, applying the chain rule, we have:
[ f'(x) = g'(u) \cdot u'(x) = 3^{-x} \ln(3) \cdot (-1) ]
So,
[ f'(x) = -3^{-x} \ln(3) ]
That's the derivative of ( f(x) = 3^{-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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