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3 years ago

Can someone help me on this point slope thing i don’t know what i’m doing

Mathematics

1 answer:

bezimeni [28]3 years ago

6 0

The equation would be y - 2 = -3(x + 9)

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11 months ago

Derivative of<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%7B3x%7D%5E%7B2%7D%20-%202x%20-%201%20%7D%7B%20%7Bx%7D%5E%7B2

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Answer:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

Step-by-step explanation:

we would like to figure out the derivative of the following:

$\displaystyle \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

to do so, let,

$\displaystyle y = \frac{ { 3x }^{2} - 2x - 1 }{ {x}^{2} }$

By simplifying we acquire:

$\displaystyle y = 3 - \frac{2}{x} - \frac{1}{ {x}^{2} }$

use law of exponent which yields:

$\displaystyle y = 3 - 2 {x}^{ - 1} - { {x}^{ - 2} }$

take derivative in both sides:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} (3 - 2 {x}^{ - 1} - { {x}^{ - 2} } )$

use sum derivation rule which yields:

$\rm\displaystyle \frac{dy}{dx} = \frac{d}{dx} 3 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

By constant derivation we acquire:

$\rm\displaystyle \frac{dy}{dx} = 0 - \frac{d}{dx} 2 {x}^{ - 1} - \frac{d}{dx} {x}^{ - 2}$

use exponent rule of derivation which yields:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ - 1 -1} ) - ( - 2 {x}^{ - 2 - 1} )$

simplify exponent:

$\rm\displaystyle \frac{dy}{dx} = 0 - ( - 2 {x}^{ -2} ) - ( - 2 {x}^{ - 3} )$

two negatives make positive so,

$\displaystyle \frac{dy}{dx} = 2 {x}^{ -2} + 2 {x}^{ - 3}$

<h3>further simplification if needed:</h3>

by law of exponent we acquire:

$\displaystyle \frac{dy}{dx} = \frac{2 }{x^2}+ \frac{2}{x^3}$

simplify addition:

$\displaystyle \frac{dy}{dx} = \frac{2x + 2}{x^3}$

and we are done!

5 0

2 years ago