# How do you calculate the derivative of (cos(t^2)+t) dt from t=-5 to t=sinx?

Jun 26, 2016

$= \cos x \left(\cos {\left(\sin x\right)}^{2} + \sin x\right)$

#### Explanation:

i think you mean this

$\frac{d}{\mathrm{dx}} {\int}_{- 5}^{\sin x} \setminus \cos {t}^{2} + t \setminus \mathrm{dt}$

We can use the Leibnitz Rule for differentiating under the integral sign which simplifies considerably here to

$\frac{d}{\mathrm{dx}} {\int}_{- 5}^{\sin x} \setminus \cos {t}^{2} + t \setminus \mathrm{dt}$

$= \cos {\left(\sin x\right)}^{2} \left(\sin x\right) ' + \sin x \left(\sin x\right) '$

$= \cos {\left(\sin x\right)}^{2} \cos x + \sin x \cos x$

$= \cos x \left(\cos {\left(\sin x\right)}^{2} + \sin x\right)$

if that seems overwhelming, we can look at the Fundamental Theorem of Calculus, specifically

if $F \left(x\right) = {\int}_{a}^{x} f \left(t\right) \setminus \mathrm{dt}$, then $F ' \left(x\right) = f \left(x\right)$

this can be extended using the chain rule to

if $F \left(x\right) = {\int}_{a}^{v \left(x\right)} f \left(t\right) \setminus \mathrm{dt}$, then $F ' \left(x\right) = f \left(v \left(x\right)\right) \cdot v ' \left(x\right)$

that's all we did here