Differentiate -5/3x: "Why is Derivative 5/3x^2?

[stat643]

i can differentiate most other simple functions.. .though can someone please help me to understand why the derivative of f(x)=-5/3x is simply 5/3x^2?

 

[Doc Al]

Do you know the power rule for differentiating functions like $x^n$? (Hint: $1/x = x^{?}$.)

 

[Kurdt]

Its basically following the rule:

$$\frac{d}{dx} ax^n = anx^{n-1}$$

where in your example $a=\frac{-5}{3}$ and $n = -1$.

EDIT: puppy interupted hence late reply

 

[stat643]

1/x = x^-1 and yes i can differentiate functions in the form x^n

 

[Doc Al]

1/x = x^-1 and yes i can differentiate functions in the form x^n

Then you should be able to solve this as that's all that's needed.

 

[stat643]

oh i see , i just didnt realize n was -1, but obviously it is wen its on the denominator, problem solved. cheers

 

[cannibal]

F(x) = $$\frac{-5}{3x}$$

The first thing you got to do is re-write the function so we get:

F(x) = $$\frac{-5}{3}$$ . $$\frac{1}{x}$$

F(x) = $$\frac{-5}{3}$$ . x$$^{-1}$$

after re-writing the function, find the most appropriated way to derivate the function in this case it would be: $$\frac{d}{dx}$$ [u$$^{n}$$] = nu$$^{n-1}$$ u'

Using this rule we get:

F(x) = $$\frac{-5}{3}$$ . x$$^{-1}$$

F'(x) = (-1) . $$\frac{-5}{3}$$ . x$$^{-2}$$ (1) , then simplify

F'(x) = $$\frac{5}{3}$$ . x$$^{-2}$$, then re-write

F'(x) = $$\frac{5}{3}$$ . $$\frac{1}{x^{2}}$$

F'(x) = $$\frac{-5}{3x^{2}}$$

 

[HallsofIvy]

You don't have to use the power rule for negative powers, you can use the quotient rule: -5/3x= f(x)/g(x) with f(x)= -5 and g(x)= 3x.
(f/g)'= (f'g- fg')(g²)

Since f'= 0 and g'= 3, that gives (-5/3x)'= ((0)(3x)- (-5)(3))/(9x²= 5/3x².

The reason I mention that is that before you can use the power rule for negative powers you have to prove it for negative powers- and you do that by using the quotient rule:
x^-n= 1/xⁿ= f/g with f(x)= 1, g(x)= xⁿ. f'= 0, g'= nx^n-1 so (x^-n)'= (1/xⁿ)'= ((0)(xⁿ)- 1(nx^n-1)/x²ⁿ= n x^n-1/x²ⁿ= n x^{(n-1)- 2n}= n x^-n-1.