Find derivatives for the functions. Assume a, b, c, and k are constants.

$f\left(t\right)=3{t}^{2}-4t+1$

Aneeka Hunt
2021-06-08
Answered

Find derivatives for the functions. Assume a, b, c, and k are constants.

$f\left(t\right)=3{t}^{2}-4t+1$

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lamanocornudaW

Answered 2021-06-09
Author has **85** answers

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

asked 2021-03-07

What is the Mixed Derivative Theorem for mixed second-order partial derivatives? How can it help in calculating partial derivatives of second and higher orders? Give examples.

asked 2022-06-22

Compute the inverse function of the following polynomial on $[0,1]$?

$f(x)=\alpha {x}^{3}-2\alpha {x}^{2}+(\alpha +1)x$

(where $\alpha $ is within $]0,3[$)

$f(x)=\alpha {x}^{3}-2\alpha {x}^{2}+(\alpha +1)x$

(where $\alpha $ is within $]0,3[$)

asked 2021-06-04

Find the derivatives of the functions.

asked 2022-06-26

Limit $\underset{x\to \mathrm{\infty}}{lim}(\mathrm{sin}\sqrt{1+x}-\mathrm{sin}\sqrt{x})$

asked 2022-06-13

Calculating $\underset{x\to 0}{lim}\frac{\mathrm{sec}(x)-1}{{x}^{2}\mathrm{sec}(x)}$

The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\mathrm{sec}(x)=\frac{1}{\mathrm{cos}(x)}$ we have

$\frac{\frac{1}{\mathrm{cos}(x)}-1}{{x}^{2}\frac{1}{\mathrm{cos}(x)}}$

And that is

$\frac{\frac{1-\mathrm{cos}(x)}{\mathrm{cos}(x)}}{\frac{{x}^{2}}{\mathrm{cos}(x)}}$

Then

$\frac{(1-\mathrm{cos}(x))\cdot \mathrm{cos}(x)}{\mathrm{cos}(x)\cdot {x}^{2}}$

And

$\frac{(1-\mathrm{cos}(x))}{{x}^{2}}$

What was wrong with my procedure?

The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\mathrm{sec}(x)=\frac{1}{\mathrm{cos}(x)}$ we have

$\frac{\frac{1}{\mathrm{cos}(x)}-1}{{x}^{2}\frac{1}{\mathrm{cos}(x)}}$

And that is

$\frac{\frac{1-\mathrm{cos}(x)}{\mathrm{cos}(x)}}{\frac{{x}^{2}}{\mathrm{cos}(x)}}$

Then

$\frac{(1-\mathrm{cos}(x))\cdot \mathrm{cos}(x)}{\mathrm{cos}(x)\cdot {x}^{2}}$

And

$\frac{(1-\mathrm{cos}(x))}{{x}^{2}}$

What was wrong with my procedure?

asked 2021-11-28

Can the functions be added, subtracted, multiplied, and divided except where the denominator is zero to
produce new functions?