negative exponents

[markraz]

how do you distribute this?

(x-a)^-1

any ideas?

I realize it can be also written as 1/(x-a)
but can the exponent actually be distributed to the terms inside the parenthesis?


thanks

 

[srmichael]

how do you distribute this?

(x-a)^-1

any ideas?

I realize it can be also written as 1/(x-a)
but can the exponent actually be distributed to the terms inside the parenthesis?


thanks

No. The exponent cannot be distributed within the parentheses. The only time you can distribute an exponent inside a parentheses is if you have either of the two following forms:

\(\displaystyle (ab)^m=a^mb^m\)

or

\(\displaystyle \left (\dfrac{a}{b} \right)^m=\dfrac{a^m}{b^m}\)

 

[markraz]

Power Rule

Hi, how do I use the power rule to find the derivative of X/(X+1)

??
any ideas?

I know how to use the quotient rule and I get

[(x+1)(1)-(x)(1)] / [(x+1)^2] = 1/(x+1)^2

 

[markraz]

Why would you want to anyway?

(2 + 3)^2 : 2^2 + 3^2 = 13, not 25

why? I want to derive
x/(x+1) or more precisely x ((x+1)^-1)

how do I derive using the power rule?
the quotient rule is fairly straight forward and the preferred method
however I'd like to see how to use the power rule

 

[pka]

Hi, how do I use the power rule to find the derivative of X/(X+1)
[(x+1)(1)-(x)(1)] / [(x+1)^2] = 1/(x+1)^2

\(\displaystyle \dfrac{x}{x+1}=x(x+1)^{-1}\)

 

[markraz]

\(\displaystyle \dfrac{x}{x+1}=x(x+1)^{-1}\)

thanks, how do I expand on this to get the actual derivative?

 

[Quaid]

I want to [calculate the derivative of]

x/(x+1)

or more precisely x (x+1)^(-1)

:idea: Expressing a ratio as a product is not more precise

how do I derive using the power rule?

Hi Mark:

We need more than the Power Rule because this ratio is not simply a power; it's a ratio of different powers.

After rewriting the ratio as a product, one could use the Product Rule. (The Power Rule will be used in one part of the Product Rule.)


Let f(x) = x

Let g(x) = (x + 1)^(-1)

The product is f(x)*g(x)


The Product Rule for derivatives says that d/dx [f(x)*g(x)] is "the derivative of the first function times the second function PLUS the derivative of the second function times the first function".

d\dx [f(x)*g(x)] = d/dx f(x) * g(x) + d/dx g(x) * f(x)

The Power Rule applies when calculating d/dx g(x)

Can you work it all out?

the quotient rule is … the preferred method

Who told you this?

Cheers!

 

[markraz]

Hi Mark:

We need more than the Power Rule because this ratio is not simply a power; it's a ratio of different powers.

After rewriting the ratio as a product, one could use the Product Rule. (The Power Rule will be used in one part of the Product Rule.)


Let f(x) = x

Let g(x) = (x + 1)^(-1)

The product is f(x)*g(x)


The Product Rule for derivatives says that d/dx [f(x)*g(x)] is "the derivative of the first function times the second function PLUS the derivative of the second function times the first function".

d\dx [f(x)*g(x)] = d/dx f(x) * g(x) + d/dx g(x) * f(x)

The Power Rule applies when calculating d/dx g(x)

Can you work it all out?




Who told you this?

Cheers!

thanks