Question

let ( f ) be the function defined by ( f(x)=int_{-1}^{5 x^{2}}left(t^{2}-t
ight)^{4} d t ). find an expression for ( f^{prime}(x) ).

Explanation:

Step1: Apply the Fundamental Theorem of Calculus and the Chain Rule

Let \(u = 5x^{2}\). By the Fundamental Theorem of Calculus, if \(F(x)=\int_{a}^{u(x)}f(t)dt\), then \(F^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)\). Here \(f(t)=(t^{2}-t)^{4}\), \(u(x) = 5x^{2}\), and \(u^{\prime}(x)=10x\).

Step2: Substitute \(u\) into \(f(t)\)

Substitute \(t = u = 5x^{2}\) into \(f(t)\). We get \(f(u(x))=( (5x^{2})^{2}-5x^{2})^{4}=(25x^{4}-5x^{2})^{4}\).

Step3: Calculate \(F^{\prime}(x)\)

Using \(F^{\prime}(x)=f(u(x))\cdot u^{\prime}(x)\), we have \(F^{\prime}(x)=(25x^{4}-5x^{2})^{4}\cdot10x\).

Answer:

\(F^{\prime}(x)=10x(25x^{4}-5x^{2})^{4}\)