(j) use the fundamental theorem of calculus to find $\frac{dy}{dx}$ for \n$y = int_{t = sqrt{ln(x)}}^{t = 4}…

Explanation:

Step1: Recall the Fundamental Theorem of Calculus and chain - rule

If $y=\int_{a}^{u(x)}f(t)dt$, then $\frac{dy}{dx}=f(u(x))\cdot u'(x)$. Here, $a = 4$, $u(x)=\sqrt{\ln x}$, and $f(t)=e^{-t^{2}}$.

Step2: Find the derivative of $u(x)$

Let $u(x)=\sqrt{\ln x}=(\ln x)^{\frac{1}{2}}$. Using the chain - rule, if $y = u^{\frac{1}{2}}$ and $u=\ln x$, then $u'(x)=\frac{1}{2}(\ln x)^{-\frac{1}{2}}\cdot\frac{1}{x}=\frac{1}{2x\sqrt{\ln x}}$.

Step3: Apply the Fundamental Theorem of Calculus

Substitute $u(x)$ into $f(t)$ and multiply by $u'(x)$. We have $f(u(x)) = e^{-(\sqrt{\ln x})^{2}}=e^{-\ln x}=\frac{1}{x}$. Then $\frac{dy}{dx}=f(u(x))\cdot u'(x)=\frac{1}{x}\cdot\frac{1}{2x\sqrt{\ln x}}$.

Answer:

$\frac{1}{2x^{2}\sqrt{\ln x}}$