+26367 f(x) = (3x-6)/x
\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& \frac{6-3x}{x^2} + \frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)
+26367 f(x) = (3x-6)/x
\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& \frac{6-3x}{x^2} + \frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)
What happened in step 3?
"+ 3 * x-1"
Im confused
I actually just multiply it out to be: 3xx-1 - 6x-1 = 3 - 6x-1
Differentiation to get 0 + 6/x2 = 6/x2
+118609 thanks Heureka,
I'll just expand on your answer a little :)
\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ &&\text{Heurka has used the product rule}\\ &&u=3x-6, \quad v=x^{-1}\\ &&u'=3, \quad v'=-1x^{-2}\\ && f'(x)=uv'+u'v \\~ \\f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& -(3x-6)x^{-2} \;\;+\;\; 3 x^{-1} \\ f'(x) &=& (6-3x)x^{-2} \;\;+\;\; 3 x^{-1} \\ f'(x) &=& \frac{(6-3x)}{x^{2}} \;\;+\;\; \frac{3}{ x} \\ f'(x) &=& \frac{6-3x}{x^2} \;\;+ \;\;\frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x}{x^2} \;\;+ \;\;\frac{3x}{x^2} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)
