Mathematics, 14.02.2021 14:00 SynysterWxlf661

Y=sqrt(x)(8x-5) find the derivative

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Answer from: aclark512

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicalsExponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$ Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$ Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$ Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$ Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$ Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$ Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$ Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

Answer from: Quest

stop cheating

step-by-step explanation:

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