# First Fundamental theorem of calculus with variable

I'm having trouble finding the derivative of the following function:

$$f(x)=x\int_{2}^{x^2}\sin(t^3)dt$$

I know I can use part 1 of the Fundamental theorem of calculus for the integral, which gives me the following result:

$$\frac{\mathrm{d} }{\mathrm{d} x} \Biggl(\int_{2}^{x^2}\sin(t^3)dt\Biggr) = \sin(x^6)2x$$

But I don't know how to include the x. I thought of maybe using the Product Rule, which would look like something like this:

$$(x)\sin(x^6)2x+ (1)*(???)$$

But finding that integral, and not jumping ahead to its derivative, is no easy task for me... Am I thinking about this wrong, or is there another way?

• Welcome to Maths SX! You're right, you have to use the product rule. – Bernard Apr 17 '18 at 21:16
• The ??? is just the integral. This is the product rule. The integral is just another function of $x$; don't let it intimidate you. – saulspatz Apr 17 '18 at 21:16

Take it step by step. As you suggested, you should write $f$ as a product: $f(x) = u(x)v(x^2)$ where $u(x)=x$ and $v(x)=\int_2^x sin(t^3)dt$.
Then you should use the fundamental theorem of calculus to get the derivative of $v(x^2)$ first and then use the product rule just as you thought.
You would have $v'(x) = sin(x^3)$ and $u'(x)=1$.
Thus $f'(x) = u'(x)v(x^2)+2x\times v'(x^2)u(x)$
In the end: $$f'(x) =\int_2^{x^2} sin(x^3)dx+2x^2sin(x^6)$$