Question

find the gradient function of:
PLEASE HELP ME BE EXPLAIN

Answer 1

\[2x - \frac{ 5 }{ x ^{2} }\]

Answer 2

umm
d/dx (x^n) = n*x^(n-1)

Answer 3

u know that formula?

Answer 4

yes i know that 2x would be just 2 but i don't know how to do the fvractiob

Answer 5

use below exponent rule : \[\large \dfrac{1}{x^n} = x^{-n}\]

Answer 6

and use the same formula

Answer 7

so it equals x

Answer 8

how ?

Answer 9

because x ^1-2 = x

Answer 10

\[\large 2x - \frac{ 5 }{ x ^{2} }~~ = ~~2x - 5x^{-2}\]

Answer 11

you need to subtract 1 from the exponent okay ?

Answer 12

taking derivative you get :

\[\large\begin{align} \dfrac{d}{dx}\left(2x - \frac{ 5 }{ x ^{2} }\right)~~& = ~~\dfrac{d}{dx}\left(2x - 5x^{-2}\right)\\~\\& = ~~\dfrac{d}{dx}\left(2x\right) -\dfrac{d}{dx}\left( 5x^{-2}\right)\end{align}\]

Answer 13

you can pull out a constant from derivative as \(\large \left(c*f(x)\right)' = c * \left(f(x)\right)'\)

Answer 14

\[\large\begin{align} \dfrac{d}{dx}\left(2x - \frac{ 5 }{ x ^{2} }\right)~~& = ~~\dfrac{d}{dx}\left(2x - 5x^{-2}\right)\\~\\& = ~~\dfrac{d}{dx}\left(2x\right) -\dfrac{d}{dx}\left( 5x^{-2}\right)\\~\\ &= ~~2\dfrac{d}{dx}\left(x\right) -5\dfrac{d}{dx}\left( x^{-2}\right)\\~\\\end{align}\]

Answer 15

so if the fraction was \[\frac{ 2}{ x }\]

Answer 16

it would equal 1/-x

Answer 17

lets finish the present problem first,
we can look into 2/x later okay ?

Answer 18

ok yes

Answer 19

i just understand the d stuff?

Answer 20

whats the derivative of `x` ?

Answer 21

which x? sorry

Answer 22

\(\large \dfrac{d}{dx}\left(x\right) = ?\)

Answer 23

\[\large\begin{align} \dfrac{d}{dx}\left(2x - \frac{ 5 }{ x ^{2} }\right)~~& = ~~\dfrac{d}{dx}\left(2x - 5x^{-2}\right)\\~\\& = ~~\dfrac{d}{dx}\left(2x\right) -\dfrac{d}{dx}\left( 5x^{-2}\right)\\~\\ &= ~~2\color{red}{\dfrac{d}{dx}\left(x\right) }-5\dfrac{d}{dx}\left( x^{-2}\right)\\~\\\end{align}\]

Answer 24

that x ^^

Answer 25

2 right

Answer 26

nope

Answer 27

use the exponent formula : x = x^1

\(\large \dfrac{d}{dx}\left(x\right) = \large \dfrac{d}{dx}\left(x^1\right) = 1.x^{1-1} = ?\)

Answer 28

1

Answer 29

very good! what about the derivative of \(\large x^{-2}\) ?

Answer 30

\(\large \dfrac{d}{dx}\left(x^{-2}\right) = ? \)

Answer 31

2x^-3

Answer 32

sure ?

Answer 33

no wait -2x^-3

Answer 34

Yes! plug them in

Answer 35

plug them into what sorry

Answer 36

\[\large\begin{align} \dfrac{d}{dx}\left(2x - \frac{ 5 }{ x ^{2} }\right)~~& = ~~\dfrac{d}{dx}\left(2x - 5x^{-2}\right)\\~\\& = ~~\dfrac{d}{dx}\left(2x\right) -\dfrac{d}{dx}\left( 5x^{-2}\right)\\~\\ &= ~~2\color{red}{\dfrac{d}{dx}\left(x\right) }-5\dfrac{d}{dx}\left( x^{-2}\right)\\~\\&= ~~2\color{red}{\left(1\right) }-5\left( -2x^{-3}\right)\\~\\\end{align}\]

Answer 37

simplify if you can

Answer 38

6x^-3?

Answer 39

ugh ?

Answer 40

thats it simplifyed

Answer 41

how ?

Answer 42

2x1x-5 = -10

Answer 43

wait no

Answer 44

\[\large\begin{align} \dfrac{d}{dx}\left(2x - \frac{ 5 }{ x ^{2} }\right)~~& = ~~\dfrac{d}{dx}\left(2x - 5x^{-2}\right)\\~\\& = ~~\dfrac{d}{dx}\left(2x\right) -\dfrac{d}{dx}\left( 5x^{-2}\right)\\~\\ &= ~~2\color{red}{\dfrac{d}{dx}\left(x\right) }-5\dfrac{d}{dx}\left( x^{-2}\right)\\~\\&= ~~2\color{red}{\left(1\right) }-5\left( -2x^{-3}\right)\\~\\&= 2 + 10x^{-3}\\~\\&=2+\dfrac{10}{x^3}\end{align}\]

Answer 45

thats the final simplified form ^^

Answer 46

what noooooo! i hate this i really don't understand :( how is the x cubed on the bottom now?

Answer 47

it got nothing to do with derivatives or calculus, so don't hate calculus yet

Answer 48

its an algebra property :
\[\large a^{-n} = \dfrac{1}{a^n}\]

Answer 49

oh so whenever its a negative power it becomes the denominator of a fraction

Answer 50

see this quick video
https://www.khanacademy.org/math/pre-algebra/exponents-radicals/negative-exponents-tutorial/v/negative-exponents