# Differentiate -5/3x

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?

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

Kurdt
Staff Emeritus
Gold Member
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 :tongue:

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

Doc Al
Mentor
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.

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

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