Question

assignment
find the derivative of y with respect to x

1. $y = \int_{0}^{\ln x} \sin e^{t} dt$
2. $y = \int_{e^{4\sqrt{x}}}^{e^{2x}} \ln t dt$

Explanation:

Step1: Recall the fundamental theorem of calculus

If $y = \int_{a}^{u(x)}f(t)dt$, then $y^\prime=f(u(x))\cdot u^\prime(x)$.

Step2: For problem 35

Let $u(x)=\ln x$, $a = 0$, and $f(t)=\sin e^{t}$. First, find $u^\prime(x)$. Since $u(x)=\ln x$, then $u^\prime(x)=\frac{1}{x}$. By the fundamental theorem of calculus, $y^\prime=\sin e^{\ln x}\cdot\frac{1}{x}$. Since $e^{\ln x}=x$, we have $y^\prime=\frac{\sin x}{x}$.

Step3: For problem 36

Let $u(x)=e^{2x}$, $a = e^{\sqrt{x}}$, and $f(t)=\ln t$. First, find $u^\prime(x)$. Since $u(x)=e^{2x}$, then $u^\prime(x)=2e^{2x}$. By the fundamental theorem of calculus, $y^\prime=\ln(e^{2x})\cdot2e^{2x}-\ln(e^{\sqrt{x}})\cdot\frac{e^{\sqrt{x}}}{2\sqrt{x}}$. Since $\ln(e^{m}) = m$, we have $y^\prime=2x\cdot2e^{2x}-\sqrt{x}\cdot\frac{e^{\sqrt{x}}}{2\sqrt{x}}=4xe^{2x}-\frac{1}{2}e^{\sqrt{x}}$.

Answer:

1. $y^\prime=\frac{\sin x}{x}$
2. $y^\prime=4xe^{2x}-\frac{1}{2}e^{\sqrt{x}}$