Chasity Kane

2022-10-08

Finding the derivative of an integral $$g(x)={\int}_{2x}^{6x}\frac{u+2}{u-4}du$$

Kaleb Harrell

Beginner2022-10-09Added 14 answers

Let $f(u)=\frac{u+2}{u-4}$ , and let F(u) be the antiderivative of f(u). Then

$${g}^{\prime}(x)=\frac{d}{dx}{\int}_{2x}^{6x}f(u)du=\frac{d}{dx}(F(u){{\textstyle |}}_{2x}^{6x})=\frac{d}{dx}[F(6x)-F(2x)]=6{F}^{\prime}(6x)-2{F}^{\prime}(2x)$$

But ${F}^{\prime}(u)=f(u)$. So the above evaluates to

$$6f(6x)-2f(2x)=6\frac{6x+2}{6x-4}-2\frac{2x+2}{2x-4}=\cdots .$$

In general,

$$\frac{d}{dx}{\int}_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot {b}^{\prime}(x)-f(a(x))\cdot {a}^{\prime}(x).$$

$${g}^{\prime}(x)=\frac{d}{dx}{\int}_{2x}^{6x}f(u)du=\frac{d}{dx}(F(u){{\textstyle |}}_{2x}^{6x})=\frac{d}{dx}[F(6x)-F(2x)]=6{F}^{\prime}(6x)-2{F}^{\prime}(2x)$$

But ${F}^{\prime}(u)=f(u)$. So the above evaluates to

$$6f(6x)-2f(2x)=6\frac{6x+2}{6x-4}-2\frac{2x+2}{2x-4}=\cdots .$$

In general,

$$\frac{d}{dx}{\int}_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot {b}^{\prime}(x)-f(a(x))\cdot {a}^{\prime}(x).$$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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