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Scrat [10]
3 years ago
12

Calculate y' using chain rule. y = (e ^ (1 / x))/(x ^ 2)

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

The answer is y^{'} =e^{\frac{1}{x} } (2x-1)

Step-by-step explanation:

First of all, we have product of 2 functions e^{1/x} and x^{2}

y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}

Let u=e^{1/x} and v=x^{2}

using chain rule for u, \frac{du}{dx}=-\frac{1}{x^2}

and \frac{dv}{dx}=2x

Therefore,

y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}=e^{1/x}(2x)+x^{2} (-\frac{1}{x^2})=e^{1/x}(2x-1)

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