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Not for me though.

 

f(x) = (3x-6)/x

Guest Jul 25, 2017

Best Answer 

 #1
avatar+18607 
+2

f(x) = (3x-6)/x

 

\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& \frac{6-3x}{x^2} + \frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)

 

laugh

heureka  Jul 25, 2017
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4+0 Answers

 #1
avatar+18607 
+2
Best Answer

f(x) = (3x-6)/x

 

\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& \frac{6-3x}{x^2} + \frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)

 

laugh

heureka  Jul 25, 2017
 #2
avatar
+1

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

Guest Jul 25, 2017
 #4
avatar+90540 
0

thanks Heureka,

I'll just expand on your answer a little :)

 

\(\begin{array}{|rcll|} \hline f(x) &=& (3x-6)/x \\ f(x) &=& (3x-6)\cdot x^{-1} \\ &&\text{Heurka has used the product rule}\\ &&u=3x-6, \quad v=x^{-1}\\ &&u'=3, \quad v'=-1x^{-2}\\ && f'(x)=uv'+u'v \\~ \\f'(x) &=& (3x-6)\cdot (-1)\cdot x^{-2} + 3 \cdot x^{-1} \\ f'(x) &=& -(3x-6)x^{-2} \;\;+\;\; 3 x^{-1} \\ f'(x) &=& (6-3x)x^{-2} \;\;+\;\; 3 x^{-1} \\ f'(x) &=& \frac{(6-3x)}{x^{2}} \;\;+\;\; \frac{3}{ x} \\ f'(x) &=& \frac{6-3x}{x^2} \;\;+ \;\;\frac{3}{x} \cdot \frac{x}{x} \\ f'(x) &=& \frac{6-3x}{x^2} \;\;+ \;\;\frac{3x}{x^2} \\ f'(x) &=& \frac{6-3x+3x}{x^2} \\ f'(x) &=& \frac{6}{x^2} \\ \hline \end{array}\)

Melody  Jul 25, 2017
 #3
avatar+90540 
+1

f(x) = (3x-6)/x

 

You could also use the quotient rule:

 

\(f(x) = \frac{(3x-6)}{x}\\ f'(x) = \frac{(x*3)-1(3x-6)}{x^2}\\ f'(x) = \frac{3x-3x+6}{x^2}\\ f'(x) = \frac{6}{x^2}\\\)

Melody  Jul 25, 2017
edited by Melody  Jul 25, 2017

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