How do you find the derivative of
y = xe^-2x?
In google type
"library wolfram step by step derivate"
then click for first term in list.
it doesn't work
Go to calc101 com
Click derivates and type
x e^(-2x)
with space betven x and e^(-2x)
and click DO IT
Using the product rule
(1)(e^-2x)+(x)(-2x)(e^-2x)
(e^-2x)(1+2x^2)
To find the derivative of y = xe^-2x, we can use the product rule of differentiation. The product rule states that if you have two functions, u(x) and v(x), then the derivative of their product, w(x) = u(x) * v(x), is given by:
w'(x) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = x and v(x) = e^-2x.
First, let's find the derivative of u(x) = x. The derivative of x with respect to x is simply 1. So u'(x) = 1.
Next, let's find the derivative of v(x) = e^-2x. To differentiate e^-2x, we can use the chain rule, which states that if we have a composition of functions, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
In this case, f(x) = e^x and g(x) = -2x.
The derivative of f(x) = e^x is simply e^x. So f'(x) = e^x.
The derivative of g(x) = -2x with respect to x is -2. So g'(x) = -2.
Now we can use the product rule to find the derivative of y = xe^-2x:
y' = u'(x) * v(x) + u(x) * v'(x)
= 1 * e^-2x + x * (-2) * e^-2x
Simplifying this expression, we get:
y' = e^-2x - 2xe^-2x
Therefore, the derivative of y = xe^-2x is y' = e^-2x - 2xe^-2x.