By the Fundamental Theorem of Calculus, we have that \(\displaystyle {f'(x) = \frac{d}{dx} \int_0^x \sin \sqrt{t^2 + \frac{\pi ^2}{4}} dt = \sin \sqrt{x^2 + \frac{\pi ^2}{4}}\). Thus, \(\displaystyle f'(0) = \sin \sqrt{0^2 + \frac{\pi ^2}{4}} =1\). 

Via the Fundamental Theorem of Calculus, we know that, given a function\(\displaystyle f(x)=\int_{0}^{x}(2t^{2}+5t-4)dt\), \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(2t^{2}+5t-4)dt=2x^{2}+5x-4\).

Therefore \(\displaystyle f'(4)=2(4)^{2}+5(4)-4=32+20-4=52-4=48\).

Via the Fundamental Theorem of Calculus, we know that, given a function \(\displaystyle f(x)=\int_{0}^{x}(6t^{2}-3t+10)dt\), \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(6t^{2}-3t+10)dt=6x^{2}-3x+10\). Therefore, \(\displaystyle f'(1)=6(1)^{2}-3(1)+10=6-3+10=3+10=13\).

Via the Fundamental Theorem of Calculus, we know that, given a function \(\displaystyle f(x)=\int_{0}^{x}(t^{2}-2t+5)dt\), \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(t^{2}-2t+5)dt=x^{2}-2x+5\). Therefore, \(\displaystyle f'(7)=(7)^{2}-2(7)+5=49-14+5=35+5=40\).

We can view the function \(\displaystyle \small g(x)\) as a function of \(\displaystyle \small \cos x\), as so

\(\displaystyle \small g(x)=F(\cos x)\)

where \(\displaystyle \small F(x)=\int_0^x (1+\sin t)dt\).

We can find the derivative of \(\displaystyle \small g(x)\) using the chain rule:

\(\displaystyle \small g'(x)=F'(\cos x)\cdot (\cos x)'=F'(\cos x)\cdot (-\sin x)\)

where \(\displaystyle \small F'(z)\) can be found using the fundamental theorem of calculus:

\(\displaystyle \small \small F'(x)=\frac{d}{dz}\int_0^z (1+\sin t)dt=1+\sin z\)

So we get

\(\displaystyle \small \small \small \small g'(x)=-\sin x\cdot F'(\cos x)=-\sin x\cdot[1+\sin(\cos x)]\)

By the Fundamental Theorem of Calculus, we know that given a function\(\displaystyle f(x)=\int_{0}^{x}(4t^{2}-6t+9)dt\), \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(4t^{2}-6t+9)dt=4x^{2}-6x+9\) .

Thus, \(\displaystyle f'(3)=4(3)^{2}-6(3)+9=36-18+9=18+9=27\)

By the Fundamental Theorem of Calculus, we know that given a function\(\displaystyle f(x)=\int_{0}^{x}(t^{2}+7t-12)dt\), \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(t^{2}+7t-12)dt=t^{2}+7t-12\) .

Thus, \(\displaystyle f'(2)=(2)^{2}+7(2)-12=4+14-12=4+2=6\)

By the Fundamental Theorem of Calculus, we know that given a function\(\displaystyle f(x)=\int_{0}^{x}(8t^{2}-t+14)dt\),  \(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}(8t^{2}-t+14)dt=8x^{2}-x+14\) .

Thus, \(\displaystyle f'(5)=8(5)^{2}-(5)+14=200-5+14=200+9=209\)

By the Fundamental Theorem of Calculus, for all functions \(\displaystyle f\) that are continuously defined on the interval \(\displaystyle [a,b]\) with \(\displaystyle x\) in \(\displaystyle [a,b]\) and for all functions \(\displaystyle F\) defined by by \(\displaystyle F(x)=\int_{a}^{x}f(t)dt\), we know that \(\displaystyle F(x)=f'(x)\).

Thus, for 

\(\displaystyle f(x)=\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt\), 

\(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{1}{t^{2}}-2t+3\right)dt=\frac{1}{x^{2}}-2x+3\).

Therefore, 

\(\displaystyle f'(6)=\frac{1}{6^{2}}-2(6)+3=\frac{1}{36}-12+3=\frac{1}{36}-9=\frac{1}{36}-\frac{324}{36}=-\frac{323}{36}\)

By the Fundamental Theorem of Calculus,  for all functions \(\displaystyle f\) that are continuously defined on the interval \(\displaystyle [a,b]\) with \(\displaystyle x\) in \(\displaystyle [a,b]\) and for all functions \(\displaystyle F\) defined by by \(\displaystyle F(x)=\int_{a}^{x}f(t)dt\), we know that \(\displaystyle F(x)=f'(x)\).  

Given 

\(\displaystyle f(x)=\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt\), then 

\(\displaystyle f'(x)=\frac{d}{dx}\int_{0}^{x}\left(\frac{2}{t^{2}+4t-5}\right)dt=\frac{2}{x^{2}+4x-5}\).

Therefore, 

\(\displaystyle f'(2)=\frac{2}{2^{2}+4(2)-5}=\frac{2}{4+8-5}=\frac{2}{4+3}=\frac{2}{7}\).