How do I differentiate $f(x) = 7 + 6/x + 6/x^2$?

Question

the problem is the following:

with the definition of the derivative, calculate

f(x) = $7+\frac 6x+ \frac6{x^2}$

I tried to solve it a bunch of times but I just don't get the correct answer

**edit: I must solve it with the def of the derivative

Answer 1

Given $7+\dfrac6x+\dfrac{6}{x^2}$

$$\dfrac{d}{dx}\left(7+\dfrac6x+\dfrac{6}{x^2}\right)$$ $$\dfrac{d}{dx}(7)=0$$ $$\dfrac{d}{dx}\left(\dfrac6x\right)=6\dfrac{d}{dx}(x^{-1})=-6x^{-2}=-\dfrac6{x^2}$$ $$\dfrac{d}{dx}\left(\dfrac{6}{x^2}\right)=6\dfrac{d}{dx}\left(x^{-2}\right)=6(-2)(x^{-3})=-\dfrac{12}{x^3}$$ $$\dfrac{d}{dx}\left(7+\dfrac6x+\dfrac{6}{x^2}\right)=-\dfrac{6}{x^2}-\dfrac{12}{x^3}$$

Answer 2

$$f’(x) = \frac{-6}{x^2} + \frac{-12}{x^3}$$

Here we are basically using the formula :

$$(\frac{u}{v} )’ = \frac{u’v-v’u}{v^2}$$

Answer 3

Given $$f(x)=\frac{6}{x^2}+\frac{6}{x}+7$$ Therefore: $$f(x)=6x^{-2}+6x^{-1}+7$$ From $\frac{d}{dx}x^n=nx^{n-1}$, we know that $$f'(x)=-2\cdot6\cdot x^{-3}-1\cdot6\cdot x^{-2}$$ Therefore $$f'(x)=-\frac{12}{x^3}-\frac{6}{x^2}$$