Question

Suppose f(x)= INT(1,x^2) ((sin(t))/t)dt. What is f'(x)?

Answer 1

D

Explanation:

Here we just need to plug in x² into the t values and then multiply by the derivative of x²

• 
  `((sin(x^(2) ))/(x^(2) ) )(2x) = (2sin(x^(2) ))/(x)`

Answer 2

`\textbf{D. }f'(x)=(2sin((x^2)))/(x)`

Explanation:

The fundamental rule of calculus tells you when ...

`f(x)=\displaystyle\int_a^u{g(t)\,dt}\\\\f'(x)=g(u)u'`

We have g(t) = sin(t)/t, and u(x) = x^2, so ...

`f'(x)=(sin((x^2)))/(x^2)(2x)=(2sin((x^2)))/(x) \qquad\text{matches D}`