How do you find the derivative of $y = x^{2} e^{- \frac{1}{x}}$ $y = x^2e^(-1/x)$ ?

Question

How do you find the derivative of $y = x^{2} e^{- \frac{1}{x}}$ $y = x^2e^(-1/x)$ ?

1 Answer

Impact of this question

9642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-04T10:04:29

$\frac{d y}{d x} = (1 + 2 x) e^{- \frac{1}{x}}$ $dy/dx=(1+2x)e^(-1/x)$ .

Explanation:

$y = x^{2} e^{- \frac{1}{x}}$ $y=x^2e^(-1/x)$

By the Product Rule, we get,

$\frac{d y}{d x} = x^{2} \cdot \frac{d}{d x} (e^{- \frac{1}{x}}) + e^{- \frac{1}{x}} \cdot \frac{d}{d x} (x^{2})$ $dy/dx=x^2*d/dx(e^(-1/x))+e^(-1/x)*d/dx(x^2)$

$= x^{2} \cdot e^{- \frac{1}{x}} \cdot \frac{d}{d x} (- \frac{1}{x}) + 2 x \cdot e^{- \frac{1}{x}}$ $=x^2*e^(-1/x)*d/dx(-1/x)+2x*e^(-1/x)$ .......[Chain Rule]

$= x^{2} \cdot e^{- \frac{1}{x}} \cdot (\frac{1}{x^{2}}) + 2 x \cdot e^{- \frac{1}{x}}$ $=x^2*e^(-1/x)*(1/x^2)+2x*e^(-1/x)$

$:. dy/dx=(1+2x)e^(-1/x)$ .

Science

Math

Humanities

... and beyond

How do you find the derivative of $y = x^{2} e^{- \frac{1}{x}}$ $y = x^2e^(-1/x)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of y = x^2e^(-1/x)y=x2e−1x y = x^2e^(-1/x)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the derivative of $y = x^{2} e^{- \frac{1}{x}}$ $y = x^2e^(-1/x)$ ?