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How do you find the derivative of #y=2^(-3/x)#?

Answer 1

Scarlett Allison

#dy/dx = 2^(-3/x)(3/x^2)ln2#

Use logarithmic differentiation.

#lny = ln(2^(-3/x))#

#lny = -3/xln(2)#

#lny = -3x^-1(ln2)#

#lny = -3ln2x^-1#

#1/y(dy/dx)= 3ln2x^(-2)#

#dy/dx = 2^(-3/x)(3/x^2)ln2#

Hopefully this helps!

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Answer 2

Emilia Aguilar

To find the derivative of ( y = 2^{-\frac{3}{x}} ), you would use the chain rule. First, rewrite the function as ( y = e^{-\frac{3}{x} \ln(2)} ). Then, differentiate it using the chain rule, which states that ( \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) ). The derivative of ( e^u ) with respect to ( x ) is ( e^u \cdot u' ). So, the derivative of ( y ) with respect to ( x ) is:

[ y' = e^{-\frac{3}{x} \ln(2)} \cdot \left(-\frac{3}{x}\right)' ]

[ = e^{-\frac{3}{x} \ln(2)} \cdot \left(-\frac{3}{x^2}\right) \cdot (\ln(2)) ]

[ = -\frac{3\ln(2)}{x^2} \cdot e^{-\frac{3}{x} \ln(2)} ]

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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.