Find the derivative of $f\left(x\right)={x}^{2}+3x$ at x. That is, find f′(x).

Travis Fogle
2021-11-19
Answered

Find the derivative of $f\left(x\right)={x}^{2}+3x$ at x. That is, find f′(x).

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Harr1957

Answered 2021-11-20
Author has **18** answers

Step 1

We have to find the derivative of f(x) at x :

$f\left(x\right)={x}^{2}+3x$

We will use following formula, for derivatives,

$\frac{df\left(x\right)}{dx}={f}^{\prime}\left(x\right)$

$\frac{da{x}^{n}}{dx}=an{x}^{n-1}$

$\frac{dx}{dx}=1$

Step 2

So finding derivative with respect to 'x', we get

$\frac{df\left(x\right)}{dx}=\frac{d({x}^{2}+3x)}{dx}$

$f}^{\prime}\left(x\right)=\frac{{dx}^{2}}{dx}+3\frac{dx}{dx$

$=2{x}^{2-1}+3\times 1$

=2x+3

Hence, value of f'(x) is 2x+3.

We have to find the derivative of f(x) at x :

We will use following formula, for derivatives,

Step 2

So finding derivative with respect to 'x', we get

=2x+3

Hence, value of f'(x) is 2x+3.

asked 2021-05-23

Use the given graph to estimate the value of each derivative.(Round all answers to one decimal place.)Graph uploaded below.

(a) f ' (0)1

(b) f ' (1)2

(c) f ' (2)3

(d) f ' (3)4

(e) f ' (4)5

(f) f ' (5)6

(a) f ' (0)1

(b) f ' (1)2

(c) f ' (2)3

(d) f ' (3)4

(e) f ' (4)5

(f) f ' (5)6

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Given is the sequence ${x}_{1}=0,\phantom{\rule{thickmathspace}{0ex}}{x}_{n+1}=\sqrt{2+{x}_{n}}$. Prove:

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

$\underset{n\to \mathrm{\infty}}{lim}{2}^{n}\sqrt{2-{x}_{n}}=\pi $

Hint:

Use the following formulas:

$\mathrm{cos}\left(\frac{x}{2}\right)=\sqrt{\frac{1+\mathrm{cos}x}{2}}$

$\mathrm{sin}\left(\frac{x}{2}\right)=\sqrt{\frac{1-\mathrm{cos}x}{2}}$

Any idea how to solve this problem?

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Let f be continuous on [a,b] and differential on (a,b). Prove that if a >= 0 there are x1, x2, x3 ∈ (a,b) such that

${f}^{\prime}({x}_{1})=(b+a)\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}=({b}^{2}+ba+{a}^{2})\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}$

I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!

${f}^{\prime}({x}_{1})=(b+a)\frac{{f}^{\prime}({x}_{2})}{2{x}_{2}}=({b}^{2}+ba+{a}^{2})\frac{{f}^{\prime}({x}_{3})}{3{x}_{3}^{2}}$

I think this problem will use the generalized mean value theorem to solve. However, I don't know how to apply it. Can you suggest a solution? Thank you very much!

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