Question

a. if ( f(x)=int_{3}^{x} \frac{1}{t} d t ), then ( f^{prime}(x)= )
b. if ( f(x)=int_{x}^{14} \frac{1}{t} d t ), then ( f^{prime}(x)= )
c. if ( f(x)=int_{13}^{x^{3}} \frac{1}{t} d t ), then ( f^{prime}(x)= )
d. if ( f(x)=int_{2+cos x}^{x^{2}+1} \frac{1}{t} d t ), then ( f^{prime}(x)= )
note: you can earn partial credit on this problem.

Explanation:

Part a: Apply Fundamental Theorem of Calculus (FTC) 1

FTC 1 states $\frac{d}{dx}\int_a^x f(t)dt = f(x)$. Here $f(t)=\frac{1}{t}$, so $F'(x)=\frac{1}{x}$.

Part b: Reverse limits + FTC 1

$\int_x^{14}\frac{1}{t}dt=-\int_{14}^x\frac{1}{t}dt$. Derivative: $-\frac{1}{x}$.

Part c: Chain rule + FTC 1

Let $u=x^3$, then $F(x)=\int_{13}^u\frac{1}{t}dt$. $F'(x)=f(u)\cdot u'=\frac{1}{x^3}\cdot 3x^2=\frac{3}{x}$.

Part d: Chain rule for both limits

$\int_{2+\cos x}^{x^2+1}\frac{1}{t}dt=\int_{a}^{x^2+1}\frac{1}{t}dt-\int_{a}^{2+\cos x}\frac{1}{t}dt$. Derivative: $\frac{1}{x^2+1}\cdot 2x - \frac{1}{2+\cos x}\cdot(-\sin x)=\frac{2x}{x^2+1}+\frac{\sin x}{2+\cos x}$.

Answer:

a. $\frac{1}{x}$
b. $-\frac{1}{x}$
c. $\frac{3}{x}$
d. $\frac{2x}{x^2 + 1} + \frac{\sin x}{2 + \cos x}$