What is the value of f'(-4)?

Expert intro: When a function is defined as an integral with variable upper limit, its derivative is given directly by the integrand. The only extra step is reading the graph value at the required x-coordinate.

Detailed walkthrough

Since

$f(x)=\int_{3}^{x} g(t)\,dt,$

by the Fundamental Theorem of Calculus,

$f'(x)=g(x).$

Therefore,

$f'(-4)=g(-4).$

To get the numeric value, read the $y$-value of the graph of $g$ at $x=-4$.

Final answer

$f'(-4)=g(-4)$

A numerical value depends on the graph of $g(x)$ at $x=-4$.

💡 Pitfall guide

A frequent mistake is to differentiate the integral limits incorrectly and try to compute an area. For $f'(x)$, you do not need the accumulated area; you only need the value of $g$ at that point.

🔄 Real-world variant

If the upper limit were $x^2$ instead of $x$, then the derivative would be $f'(x)=g(x^2)\cdot 2x$ by the chain rule. Here, with upper limit $x$, the derivative is simply $g(x)$.

🔍 Related terms

Fundamental Theorem of Calculus, derivative of an integral, function value from graph