Question

16 mark for review for ( x > 0 ), ( \frac{d}{dx} int_{1}^{sqrt{x}} \frac{1}{1 + t^2} dt = )

Explanation:

Step1: Recall Leibniz Rule

For $\frac{d}{dx}\int_{a(x)}^{b(x)} f(t)dt = f(b(x))\cdot b'(x) - f(a(x))\cdot a'(x)$

Step2: Define values

Here, $f(t)=\frac{1}{1+t^2}$, $b(x)=\sqrt{x}$, $a(x)=1$

Step3: Compute $f(b(x))$ and $b'(x)$

$f(b(x))=\frac{1}{1+(\sqrt{x})^2}=\frac{1}{1+x}$
$b'(x)=\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$

Step4: Compute $f(a(x))$ and $a'(x)$

$f(a(x))=\frac{1}{1+1^2}=\frac{1}{2}$, $a'(x)=\frac{d}{dx}(1)=0$

Step5: Apply Leibniz Rule

$\frac{d}{dx}\int_{1}^{\sqrt{x}} \frac{1}{1+t^2}dt = \frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}} - \frac{1}{2}\cdot0$

Answer:

$\frac{1}{2\sqrt{x}(1+x)}$