in exercises 39 - 42, find $dy/dx$.\n39. $y=int_{2}^{x}sqrt{2+cos^{3}t}dt$\n40. $y=int_{2}^{7x^{2}}sqrt{2+cos…

Explanation:

Step1: Recall the fundamental theorem of calculus and chain - rule

If $y=\int_{a}^{u(x)}f(t)dt$, then $\frac{dy}{dx}=f(u(x))\cdot u'(x)$. If $y = \int_{u(x)}^{v(x)}f(t)dt=\int_{a}^{v(x)}f(t)dt-\int_{a}^{u(x)}f(t)dt$, then $\frac{dy}{dx}=f(v(x))\cdot v'(x)-f(u(x))\cdot u'(x)$.

Step2: Solve for $y = \int_{2}^{7x^{2}}\sqrt{2+\cos^{3}t}dt$

Let $u = 7x^{2}$, $a = 2$ and $f(t)=\sqrt{2+\cos^{3}t}$. By the fundamental theorem of calculus and chain - rule, $\frac{dy}{dx}=f(7x^{2})\cdot(7x^{2})'$. Since $(7x^{2})' = 14x$ and $f(7x^{2})=\sqrt{2+\cos^{3}(7x^{2})}$, we have $\frac{dy}{dx}=14x\sqrt{2+\cos^{3}(7x^{2})}$.

Step3: Solve for $y=\int_{x}^{1}\frac{6}{3 + t^{4}}dt=-\int_{1}^{x}\frac{6}{3 + t^{4}}dt$

Let $u(x)=x$, $a = 1$ and $f(t)=\frac{6}{3 + t^{4}}$. By the fundamental theorem of calculus, $\frac{dy}{dx}=-f(x)\cdot(x)'$. Since $(x)' = 1$ and $f(x)=\frac{6}{3 + x^{4}}$, we have $\frac{dy}{dx}=-\frac{6}{3 + x^{4}}$.

Step4: Solve for $y=\int_{x}^{2x}\frac{1}{t^{2}+1}dt=\int_{a}^{2x}\frac{1}{t^{2}+1}dt-\int_{a}^{x}\frac{1}{t^{2}+1}dt$

Let $u(x)=x$, $v(x)=2x$ and $f(t)=\frac{1}{t^{2}+1}$. By the fundamental theorem of calculus and chain - rule, $\frac{dy}{dx}=f(2x)\cdot(2x)'-f(x)\cdot(x)'$. Since $(2x)' = 2$ and $(x)' = 1$, $f(2x)=\frac{1}{(2x)^{2}+1}=\frac{1}{4x^{2}+1}$ and $f(x)=\frac{1}{x^{2}+1}$, we have $\frac{dy}{dx}=\frac{2}{4x^{2}+1}-\frac{1}{x^{2}+1}$.

Answer:

For $y = \int_{2}^{7x^{2}}\sqrt{2+\cos^{3}t}dt$, $\frac{dy}{dx}=14x\sqrt{2+\cos^{3}(7x^{2})}$; for $y=\int_{x}^{1}\frac{6}{3 + t^{4}}dt$, $\frac{dy}{dx}=-\frac{6}{3 + x^{4}}$; for $y=\int_{x}^{2x}\frac{1}{t^{2}+1}dt$, $\frac{dy}{dx}=\frac{2}{4x^{2}+1}-\frac{1}{x^{2}+1}$