Derivative of x·e^x
Both factors depend on x. Apply the product rule to combine their two contributions to the slope.
Step-by-step solution
Let u=x and v=exp(x). Then u′=1 and v′=exp(x).
f′=u′v+uv′=exp(x)+x·exp(x).
f′(x)=(x+1)exp(x). This is equivalent to the two-term answer.
Domain and tangent behavior
The function and its derivatives exist for all real x. Because exp(x)>0, f′ changes from negative to positive at x=−1. The global minimum is f(−1)=−1/e. At x=0 the point is (0,0), with slope 1 and tangent y=x.
Second derivative
The second derivative is (x+2)exp(x). It changes sign at x=−2, giving an inflection point (−2,−2/e²). A stationary point and an inflection point need not occur at the same input.
Common mistake
Do not multiply the derivatives of the factors. That would give exp(x) and omit the x·exp(x) term.
Practice problems
Differentiate x·exp(2x)
Answer: (1+2x)exp(2x): combine the product rule and chain rule.
Differentiate x²·exp(x)
Answer: (2x+x²)exp(x): differentiate each factor once in separate terms.
Related exponential derivatives
Review the chain rule and product rule, or read our verification method and limitations.