Question

if $f(x) = \int_{3}^{x^2} t^4 dt$ then $f(x) = $ question help: video message instructor submit question question 2 0/1 pt 1 99 details if $f(u) = \int_{0}^{u^3} \sqrt{8 + z^5} dz$ then $f(u) = $ question help: video message instructor submit question question 3 0/1 pt 3 99 details if $y(x) = \int_{3}^{e^x} \frac{\cos s^9}{s^2} ds$ then $y(x) = $

Explanation:

Step1: Apply Fundamental Theorem of Calculus (FTC) + Chain Rule

For $f(x)=\int_{3}^{x^2} t^4 dt$, let $u=x^2$, so $f(x)=\int_{3}^{u} t^4 dt$. By FTC, $\frac{d}{du}\int_{3}^{u} t^4 dt = u^4$. By Chain Rule, $f'(x)=\frac{d}{du}\int_{3}^{u} t^4 dt \cdot \frac{du}{dx}$.
$\frac{du}{dx}=2x$, substitute $u=x^2$:
$f'(x)=(x^2)^4 \cdot 2x = x^8 \cdot 2x = 2x^9$

Step2: Apply FTC + Chain Rule to $f(u)$

For $f(u)=\int_{0}^{u^3} \sqrt{8+z^5} dz$, let $v=u^3$, so $f(u)=\int_{0}^{v} \sqrt{8+z^5} dz$. By FTC, $\frac{d}{dv}\int_{0}^{v} \sqrt{8+z^5} dz = \sqrt{8+v^5}$. By Chain Rule, $f'(u)=\frac{d}{dv}\int_{0}^{v} \sqrt{8+z^5} dz \cdot \frac{dv}{du}$.
$\frac{dv}{du}=3u^2$, substitute $v=u^3$:
$f'(u)=\sqrt{8+(u^3)^5} \cdot 3u^2 = 3u^2\sqrt{8+u^{15}}$

Step3: Apply FTC + Chain Rule to $y(x)$

For $y(x)=\int_{3}^{e^x} \frac{\cos s^9}{s^2} ds$, let $w=e^x$, so $y(x)=\int_{3}^{w} \frac{\cos s^9}{s^2} ds$. By FTC, $\frac{d}{dw}\int_{3}^{w} \frac{\cos s^9}{s^2} ds = \frac{\cos w^9}{w^2}$. By Chain Rule, $y'(x)=\frac{d}{dw}\int_{3}^{w} \frac{\cos s^9}{s^2} ds \cdot \frac{dw}{dx}$.
$\frac{dw}{dx}=e^x$, substitute $w=e^x$:
$y'(x)=\frac{\cos (e^x)^9}{(e^x)^2} \cdot e^x = \frac{\cos e^{9x}}{e^{2x}} \cdot e^x = \frac{\cos e^{9x}}{e^x}$

Answer:

1. $f'(x)=2x^9$
2. $f'(u)=3u^2\sqrt{8+u^{15}}$
3. $y'(x)=\frac{\cos e^{9x}}{e^x}$