What is the derivative of #x(6^(-2x))#?

Answer 1

#=(1-2xln 6)6^(-2x)#

#(x cdot 6^(-2x))'#

By Product Rule,

#=1 cdot 6^(-2x)+x cdot (6^(-2x))'#
By #(b^x)'=(ln b)b^x# and Chain Rule,
#=6^(-2x)+x cdot (ln 6)6^(-2x)(-2)#
By factoring out #6^(-2x)#,
#=(1-2xln 6)6^(-2x)#

I hope that this was clear.

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Answer 2

To find the derivative of ( x \cdot 6^{-2x} ), you can use the product rule and the chain rule of differentiation. The derivative is:

[ \frac{d}{dx}(x \cdot 6^{-2x}) = x \cdot \frac{d}{dx}(6^{-2x}) + 6^{-2x} \cdot \frac{d}{dx}(x) ]

Applying the chain rule and the power rule, the derivative of ( 6^{-2x} ) is:

[ \frac{d}{dx}(6^{-2x}) = -2 \cdot 6^{-2x} \cdot \ln(6) ]

The derivative of ( x ) is simply ( 1 ).

Substituting these derivatives back into the original expression, you get:

[ \frac{d}{dx}(x \cdot 6^{-2x}) = x \cdot (-2 \cdot 6^{-2x} \cdot \ln(6)) + 6^{-2x} \cdot 1 ]

This can be simplified as:

[ \frac{d}{dx}(x \cdot 6^{-2x}) = -2x \cdot 6^{-2x} \cdot \ln(6) + 6^{-2x} ]

So, the derivative of ( x \cdot 6^{-2x} ) is ( -2x \cdot 6^{-2x} \cdot \ln(6) + 6^{-2x} ).

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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.

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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