# What is the derivative of int_(cosx)^(7x) \ cost^3 \ dt  wrt x?

Jul 25, 2017

$\frac{d}{\mathrm{dx}} {\int}_{\cos x}^{7 x} \setminus \cos {t}^{3} \setminus \mathrm{dt} = 7 \cos {\left(7 x\right)}^{3} + \sin x \cos \left({\cos}^{3} x\right)$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$ for any constant $a$

(ie the derivative of an integral gives us the original function back).

We are asked to find:

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \setminus {\int}_{\cos x}^{7 x} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

Note that neither the upper or lower bounds are in the correct format for the FTOC to be applied, directly. We can manipulate the definite integral as follows:

${\int}_{\cos x}^{7 x} \setminus \cos {t}^{3} \setminus \mathrm{dt} = {\int}_{a}^{7 x} \setminus \cos {t}^{3} \setminus \mathrm{dt} - {\int}_{a}^{\cos x} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

And so:

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{a}^{7 x} \setminus \cos {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} {\int}_{a}^{\cos x} \setminus \cos {t}^{3} \setminus \mathrm{dt}$ ... [A]

We have arbitrary chosen the lower limit as some arbitrary constant $a$ (we could use $0$ wlog, any number will do!). We can further manipulate these definite integral using a substitution and the chain rule. Let:

$u = 7 x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 7$
$v = \cos x \implies \frac{\mathrm{dv}}{\mathrm{dx}} = - \sin x$

Then substituting into [A], and applying the chain rule, we get:

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{a}^{u} \setminus \cos {t}^{3} \setminus \mathrm{dt} - \frac{d}{\mathrm{dx}} {\int}_{a}^{v} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

$\text{ } = \frac{\mathrm{du}}{\mathrm{dx}} \frac{d}{\mathrm{du}} {\int}_{a}^{u} \setminus \cos {t}^{3} \setminus \mathrm{dt} - \frac{\mathrm{dv}}{\mathrm{dx}} \frac{d}{\mathrm{dv}} {\int}_{a}^{v} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

$\text{ } = 7 \frac{d}{\mathrm{du}} {\int}_{a}^{u} \setminus \cos {t}^{3} \setminus \mathrm{dt} - \left(- \sin x\right) \frac{d}{\mathrm{dv}} {\int}_{a}^{v} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

$\text{ } = 7 \frac{d}{\mathrm{du}} {\int}_{a}^{u} \setminus \cos {t}^{3} \setminus \mathrm{dt} + \sin x \frac{d}{\mathrm{dv}} {\int}_{a}^{v} \setminus \cos {t}^{3} \setminus \mathrm{dt}$

And now the derivative of both the integrals are in the correct form for the FTOC to be applied, giving:

$g ' \left(x\right) = 7 \cos {u}^{3} + \sin x \cos {v}^{3}$

And restoring the initial substitution we get:

$g ' \left(x\right) = 7 \cos {\left(7 x\right)}^{3} + \sin x \cos \left({\cos}^{3} x\right)$