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3 years ago

experts, geniuses, aces and moderators .. need help on the attached. will give brainliest!!! find the derivative of e^x

Mathematics

1 answer:

STatiana [176]3 years ago

4 0

To find the derivative, you must use the chain rule.

If u=x^3+2x:
dy/dx=(dy/du)(du/dx)
dy/du=d/du(e^u)=e^u=e^(x^3 + 2x)
du/dx =d/dx (x^3+2x) = 3x^2 + 2

So dy/dx=
e^(x^3+2x) * (3x^2+ 2)

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Use the Integral Test to determine whether the series is convergent or divergent

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Answer:

A. $\sum_{n=1}^{\infty}\frac{n}{e^{15n}}$ converges by integral test

Step-by-step explanation:

A. At first we need to verify that the function which the series is related ( $\frac{n}{e^{15n}}$ ) fills the necessary conditions to ensure that the test is effective.

*f(x) must be continuous or differentiable

*f(x) must be positive and decreasing

Let´s verify that $f(x)=\frac{n}{e^{15n}}$ fills these conditions:

*Considering that eˣ≠0 for all x, the function $f(x)=\frac{n}{e^{15n}}$ does not have any discontinuities, so it´s continuous

*Because eˣ is increasing:

if a<b ,then eᵃ<eᵇ

if 0<eᵃ<eᵇ ,then 1/eᵃ > 1/eᵇ

if 1/eᵃ > 1/eᵇ and a<b, then a/eᵃ<b/eᵇ

We conclude that $f(x)=\frac{n}{e^{15n}}$ is decreasing

*Because eˣ is always positive and the sum is going from 1 to ∞, this show that $f(x)=\frac{n}{e^{15n}}$ is positive in [1,∞).

Now we are able to use the integral test in $f(x)=\frac{n}{e^{15n}}$ as follows:

$\sum_{n=1}^{\infty}\frac{n}{e^{15n}}\ converges\ \leftrightarrow\ \int_{1}^{\infty}\frac{x}{e^{15x}}\ dx\ converges$

Let´s proceed to integrate f(x) using integration by parts

$\int_{1}^{\infty}\frac{x}{e^{15x}}\ dx=\int_{1}^{\infty}xe^{-15x}\ dx$

Choose your U and dV like this:

$U=x\ \rightarrow dU=1\\ dV=e^{-15x}\ \rightarrow V=\frac{-e^{-15x}}{15}$

And continue using the formula for integration by parts:

$\int_{1}^{\infty}Udv = UV|_{1}^{\infty} - \int_{1}^{\infty}Vdu$

$\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15} \int_{1}^{\infty}e^{-15x}\ dx$

$\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{-1}{15}(\frac{-1}{15e^{15x}})|_{1}^{\infty}$

$\int_{1}^{\infty}xe^{-15x}\ dx= \frac{-x}{15e^{15x}}|_{1}^{\infty} -\frac{1}{225e^{15x}}|_{1}^{\infty}$

Because we are dealing with ∞, we´d rewrite it as a limit that will help us at the end of the integral:

$\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}}(\frac{-x}{15e^{15x}}|_{1}^{b}-\frac{1}{225e^{15x}}|_{1}^{b})$

$\int_{1}^{\infty}xe^{-15x}\ dx= \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}}-(\frac{-1}{15e^{15}}-\frac{1}{225e^{15}})$

$\int_{1}^{\infty}xe^{-15x}\ dx= ( \lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}})+\frac{1}{15e^{15}}(1-\frac{1}{15})$

We only have left to solve the limits, but because b goes to ∞ and it is in an exponential function on the denominator everything goes to 0

$\lim_{b \to{\infty}} \frac{-b}{15e^{15b}}-\frac{1}{225e^{15b}} = 0$

$\int_{1}^{\infty}xe^{-15x}\ dx= \frac{1}{15e^{15}}(1-\frac{1}{15})$

Showing that the integral converges, it´s the same as showing that the series converges.

By the integral test $\sum_{n=1}^{\infty}\frac{n}{e^{15n}}$ converges

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation

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