How do you find the derivative of #(e^(2x))/(4^x)#?

Answer 1

#(e^(2x)(1-ln(2)))/(2^(2x-1))#

After the Quotient rule

#(u/v)'=(u'v-uv')/v^2#

we get

#f'(x)=(e^(2x)2*4^x-e^(2x)*4^x*ln(4))/(4^x)^2# simplifying we get
#(e^(2x)(1-ln(2)))/(2^(4x-2x-1))# and this is equal to
#(e^(2x)(1-ln(2)))/(2^(2x-1))#
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Answer 2

To find the derivative of ( \frac{e^{2x}}{4^x} ), you can use the quotient rule. The quotient rule states that for functions ( u(x) ) and ( v(x) ), the derivative of ( \frac{u(x)}{v(x)} ) is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

Applying the quotient rule to ( \frac{e^{2x}}{4^x} ):

[ u(x) = e^{2x}, \quad v(x) = 4^x ]

[ u'(x) = 2e^{2x}, \quad v'(x) = (\ln 4) \cdot 4^x ]

[ \frac{d}{dx} \left( \frac{e^{2x}}{4^x} \right) = \frac{(2e^{2x})(4^x) - (e^{2x})(\ln 4)(4^x)}{(4^x)^2} ]

[ = \frac{2e^{2x} \cdot 4^x - e^{2x} \cdot \ln 4 \cdot 4^x}{(4^x)^2} ]

[ = \frac{2e^{2x} \cdot 4^x - e^{2x} \cdot \ln 4 \cdot 4^x}{16^x} ]

[ = \frac{2e^{2x} - e^{2x} \cdot \ln 4}{4^x} ]

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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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