# How do you calculate the derivative of intsin^3(2t-1) dt  from [x, x-1]?

Jul 6, 2015

Use the Fundamental Theorem of Calculus Part 1 and some rewriting and the chain rule. The derivative is ${\sin}^{3} \left(2 x - 3\right) - {\sin}^{3} \left(2 x - 1\right)$

#### Explanation:

The Fundamental Theorem of Calculus, Part 1 (FTC 1) tells us that for

$g \left(x\right) = {\int}_{a}^{x} f \left(t\right) \mathrm{dt}$, for $x$ in some interval $\left[a , b\right]$ on which $f$ is continuous, then we have (among other things):

$g ' \left(x\right) = f \left(x\right)$

To find the derivative of

$g \left(x\right) = {\int}_{x}^{x - 1} f \left(t\right) \mathrm{dt}$, where $f \left(t\right) = {\sin}^{3} \left(2 t - 1\right)$

we note that the function is continuous on $\mathbb{R}$, so that is not a concern. But we need a constant as the lower limit of integration.

So we will rewrite.

$g \left(x\right) = {\int}_{x}^{x - 1} f \left(t\right) \mathrm{dt} = {\int}_{x}^{0} f \left(t\right) \mathrm{dt} + {\int}_{0}^{x - 1} f \left(t\right) \mathrm{dt}$

See properties of the definite integral for

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} = {\int}_{a}^{c} f \left(t\right) \mathrm{dt} + {\int}_{c}^{b} f \left(t\right) \mathrm{dt}$

(Pick any constant you like where I have used $0$.)

$g \left(x\right) = - {\int}_{0}^{x} f \left(t\right) \mathrm{dt} + {\int}_{0}^{x - 1} f \left(t\right) \mathrm{dt}$

See properties of the definite integral for

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} = - {\int}_{b}^{a} f \left(t\right) \mathrm{dt}$

We are almost there, but the second integral doesn't just have an $x$ as the upper limit of integration, so we need the chain rule version of FTC 1.

For $g \left(x\right) = {\int}_{a}^{u} f \left(t\right) \mathrm{dt}$, we get $g ' \left(x\right) = f \left(u\right) \frac{\mathrm{du}}{\mathrm{dx}}$

So for this problem we arrive at:

$g ' \left(x\right) = - f \left(x\right) + f \left(x - 1\right) \cdot 1$ (Of course, the derivative of $x - 1$ is just $1$)

So the derivative (with respect to $x$) of

${\int}_{x}^{x - 1} {\sin}^{3} \left(2 t - 1\right) \mathrm{dt}$ is:

$- {\sin}^{3} \left(2 x - 1\right) + {\sin}^{3} \left(2 \left(x - 1\right) - 1\right)$

The answer can be rewritten as:

${\sin}^{3} \left(2 x - 3\right) - {\sin}^{3} \left(2 x - 1\right)$

Final Note

There are people around who use the FTC 1 in the form:

The derivative of $g \left(x\right) = {\int}_{f \left(x\right)}^{g \left(x\right)} h \left(t\right) \mathrm{dt}$ is

$h \left(g \left(x\right)\right) \cdot g ' \left(x\right) - h \left(f \left(x\right)\right) \cdot f ' \left(x\right)$.

Memorized this way, we can skip all the set-up work I did.

I've never taught from a textbook in the US that does it this way.