what would the derivative of xe^(1/x^2) be?
i got x(xe^-2(x^(-2))-1)) + e ^x^(-2)
steps too please.
Thank you
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differentiate xe^(1/x^2)
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http://www.wolframalpha.com/input/?i=differentiate+xe^%281%2Fx^2%29
thank you
To find the derivative of the function f(x) = xe^(1/x^2), we can use the product rule and chain rule.
Let's break down the steps:
Step 1: Write the given function:
f(x) = xe^(1/x^2)
Step 2: Apply the product rule:
The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:
(fg)'(x) = f'(x)g(x) + f(x)g'(x)
In our case, u(x) = x and v(x) = e^(1/x^2). Applying the product rule:
f'(x) = (u'(x)v(x)) + (u(x)v'(x))
Step 3: Compute the derivatives:
Let's compute the derivatives separately.
u(x) = x ⇒ u'(x) = 1 (derivative of x with respect to x)
v(x) = e^(1/x^2) ⇒ v'(x) = (d/dx)(e^(1/x^2))
To find v'(x), we need to apply the chain rule, which states that if we have a function of the form g(h(x)), then its derivative is given by (g(h(x)))' = g'(h(x)) * h'(x).
In our case, g(h(x)) = e^x and h(x) = 1/x^2.
Let's calculate the derivatives needed for v'(x):
g(h) = e^x ⇒ g'(x) = e^x (derivative of e^x with respect to x)
h(x) = 1/x^2 ⇒ h'(x) = (d/dx)(1/x^2)
To find h'(x), we need to apply the chain rule again:
h(x) = 1/x^2
⇒ h(x) = x^(-2)
⇒ h'(x) = d(x^(-2))/dx
Differentiating x^(-2):
h'(x) = -2x^(-3) = -2/x^3
Now we can go back to finding v'(x) using the chain rule formula:
v'(x) = g'(h(x)) * h'(x)
= e^x * (-2/x^3)
= -2e^x/x^3
Now let's substitute the derivatives we found into the product rule formula:
f'(x) = (u'(x)v(x)) + (u(x)v'(x))
= (1 * e^(1/x^2)) + (x * -2e^x/x^3)
= e^(1/x^2) - 2xe^x/x^3
So, the derivative of f(x) = xe^(1/x^2) is e^(1/x^2) - (2xe^x)/x^3.
It seems that your answer has minor errors. However, now you have the correct solution and can compare it to yours.