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.
Why the factored answer is useful
The expanded derivative e^x+x e^x shows the two product-rule contributions clearly. Factoring gives (x+1)e^x, which makes sign analysis easier. Since e^x is positive for every real x, the derivative has the same sign as x+1. It is negative for x<−1, zero at −1, and positive for x>−1. This immediately identifies the global minimum without a sign chart for both factors.
Check the slope at specific points
At x=0, the derivative formula gives (0+1)e^0=1. A small centered secant through f(x)=xe^x also approaches 1, which matches the tangent y=x at the origin. At x=1, the slope is 2e. At x=−1, the slope is zero even though the function value is −1/e. Keeping slope and height separate prevents a common interpretation mistake.
A repeated-derivative pattern
Repeated differentiation reveals a clean pattern: f′=(x+1)e^x, f″=(x+2)e^x, and f‴=(x+3)e^x. In general, the nth derivative is (x+n)e^x. Each product-rule application differentiates the linear factor to 1 and reproduces the exponential. This pattern is useful for Taylor polynomials and for checking higher-order calculator output.
More worked variations
x·e^(2x): f′=e^(2x)+2xe^(2x)=(1+2x)e^(2x). The factor 2 comes from the chain rule.
x²e^x: f′=2xe^x+x²e^x=x(x+2)e^x. Differentiate only one factor in each product-rule term.
(x+3)e^x: f′=e^x+(x+3)e^x=(x+4)e^x. The constant shift changes the linear factor but not the exponential derivative.
Connection to integration
The product also appears when integrating xe^x. Reversing the product rule through integration by parts gives ∫xe^x dx=(x−1)e^x+C. Differentiating that antiderivative returns xe^x, providing a second independent check on the related algebra.
Questions students ask
Why is the answer not simply e^x?
That result differentiates both factors and multiplies their derivatives. The product rule must also preserve each original factor in a separate term.
Are e^x(1+x) and (x+1)e^x different?
No. Commutativity of addition and multiplication makes them identical.
Related exponential derivatives
Review the chain rule and product rule, or read our verification method and limitations.