Question

Find \(F^{\prime}(x)\). $$ F(x)=\int_{2}^{x^{2}} \frac{1}{t^{3}} d t $$

Short answer

\(F'(x) = 2/x^5\)

Step-by-step solution

Identify the inner function

The integral given is an example of a function where we need to use the Chain rule. Thus, we need to identify the inner function. Here it is \(g(x) = x^2\).

Evaluate the integral

According to the Fundamental Theorem of Calculus, the derivative of the function \(F(x)\) is the function being integrated, up to the inner function \(g(x)\). Here it is \(f(t) = 1/{t^3}\). Therefore, \(f(g(x)) = 1/{x^2^3} = 1/{x^6}\).

Apply the Chain rule

After identifying the inner function and evaluating the integral, use the Chain rule: \( (f(g(x)))' = f'(g(x))*g'(x) \). The derivative of \(g(x) = x^2\) is \(g'(x)=2x\). Multiply this with the result from Step 2 to obtain the derivative \(F'(x) = 1/{x^6} * 2x = 2/{x^5}\).