Question

assignment
find the derivative of y with respect to x

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

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, $y^\prime=\sin e^{\ln x}\cdot\frac{1}{x}$. Simplify $\sin e^{\ln x}$ to $\sin x$. So $y^\prime=\frac{\sin x}{x}$.

Step3: For problem 36

Let $u(x)=e^{2x}$, $v(x)=e^{4\sqrt{x}}$ and $f(t)=\ln t$. Then $y=\int_{v(x)}^{u(x)}f(t)dt=\int_{a}^{u(x)}f(t)dt-\int_{a}^{v(x)}f(t)dt$.
Find $u^\prime(x)$ and $v^\prime(x)$. $u^\prime(x) = 2e^{2x}$ and $v^\prime(x)=e^{4\sqrt{x}}\cdot\frac{4}{2\sqrt{x}}=\frac{2e^{4\sqrt{x}}}{\sqrt{x}}$.
By the fundamental theorem of calculus, $y^\prime=f(u(x))\cdot u^\prime(x)-f(v(x))\cdot v^\prime(x)$. Substitute $u(x),v(x),u^\prime(x),v^\prime(x)$ and $f(t)$: $y^\prime=\ln(e^{2x})\cdot2e^{2x}-\ln(e^{4\sqrt{x}})\cdot\frac{2e^{4\sqrt{x}}}{\sqrt{x}}$. Since $\ln(e^{2x}) = 2x$ and $\ln(e^{4\sqrt{x}})=4\sqrt{x}$, we have $y^\prime=2x\cdot2e^{2x}-4\sqrt{x}\cdot\frac{2e^{4\sqrt{x}}}{\sqrt{x}}=4xe^{2x}-8e^{4\sqrt{x}}$.

Answer:

1. $y^\prime=\frac{\sin x}{x}$
2. $y^\prime=4xe^{2x}-8e^{4\sqrt{x}}$