Question: what is the derivative of f(x)=x^2*e^x
Would the answer simply be 2x+e^x?
NO. f= uv
f'= u v' + v u'
here, that translates to
f'= 2xe^x+ x^2e^x
check my thinking.
can that equation be simplified?
(x+2)x e^x ??? not much simpler
To find the derivative of the function f(x) = x^2 * e^x, we can use the product rule combined with the chain rule.
The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by:
(d/dx) [u(x) * v(x)] = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = x^2 and v(x) = e^x. To find the derivative of f(x), we need to calculate the derivatives of u(x) and v(x) and then apply the product rule as mentioned above.
The derivative of u(x) = x^2 can be found using the power rule, which states that if we have a function of the form u(x) = x^n, then its derivative is given by:
(d/dx) [x^n] = n * x^(n-1)
Applying the power rule to u(x) = x^2, we get:
u'(x) = d/dx (x^2) = 2 * x^(2-1) = 2 * x^1 = 2x
The derivative of v(x) = e^x can be found using the chain rule, which states that if we have a function of the form v(x) = f(g(x)), then its derivative is given by:
(d/dx) [f(g(x))] = f'(g(x)) * g'(x)
In this case, f(x) = e^x, and g(x) = x. The derivative of f(x) = e^x is simply f'(x) = e^x, and the derivative of g(x) = x is g'(x) = 1. Therefore, applying the chain rule, we get:
v'(x) = f'(g(x)) * g'(x) = e^x * 1 = e^x
Now, we can apply the product rule to find the derivative of f(x):
f'(x) = u'(x) * v(x) + u(x) * v'(x) = 2x * e^x + x^2 * e^x = 2xe^x + x^2e^x
So, the correct derivative of f(x) = x^2 * e^x is indeed 2xe^x + x^2e^x.