3

First I set up the difference quotient as follows: $$\dfrac{6(x+h)^2-\frac{6}{x+h}-(6x^2-\frac{6}{x})}{h}$$ Then I expand the binomial in the first term, and distribute $6$ and $-1$: $$\dfrac{6x^2+12xh+6h^2-\frac{6}{x+h}-6x^2+\frac{6}{x}}{h}$$ Finally I cancel like terms: $$\dfrac{12xh+6h^2-\frac{6}{x+h}+\frac{6}{x}}{h}$$ At this point I multiply the numerator by $\frac{1}{h}$ and I'm left with: $$12x+6h-\frac{6}{h(x+h)}+\frac{6}{xh}$$ By now I grow weary because the solution is $$f'(x)=12x+\frac{6}{x^2}$$ The path I took looks unpromisng. I'm not sure where I went wrong. Taking any help I can get. Thank you.

xvymnp
  • 33
  • @JohnOmielan I find a common denominator and I'm left with $-\frac{x}{x(x+h)}+\frac{x+h}{x(x+h)}=\frac{h}{x(x+h)}$ which to my understanding approaches $0$. – xvymnp Mar 21 '24 at 05:25
  • Your problem is with figuring the derivative of $\frac 1x$ as $\lim \frac {\frac 1{x+h} -\frac 1x}h=\lim \frac 1{h(x+h)} -\frac 1{xh}$ and getting stuck. You need to put them over a common denominator. $\lim \frac {x - (x+h)}{h(x+h)x}=\lim \frac {-h}{h(x+h)x}=\lim \frac{-1}{(x+h)x} =\frac {-1}{x^2}$. (Although I'd advice putting over common divisor before you did the multiplying by $\frac 1h$) – fleablood Mar 21 '24 at 05:25
  • @xvymnp That's good progress. However, note there's a multiplier of $\frac{6}{h}$ in front, so the $h$ factors cancel. We therefore get $-\frac{6}{h(x+h)}+\frac{6}{xh}=\frac{6}{h}\left(-\frac{1}{x+h}+\frac{1}{x}\right)=\frac{6}{h}\left(\frac{h}{x(x+h)}\right)=\frac{6}{x(x+h)}$ (as shown in Dan's answer). – John Omielan Mar 21 '24 at 05:27
  • 2
    @JohnOmielan It always amazes me how the answer is right in front of my face and yet I miss it. It's clear as day now. Thank you. – xvymnp Mar 21 '24 at 05:34
  • Alternatively you can multiply top and bottom by $\frac 1h$ again and get $\frac 6{h(x+h)} -\frac 6{hx} = \frac {\frac 6h}{x+h}-\frac {\frac 6h}x$ and put over common denominator: $\frac {\frac 6h\cdot x - \frac 6h(x+h)}{x(h+x)}=\frac {\frac 6hx -\frac 6hx - \frac 6h h}{x^2+hx}= \frac {-6}{x^2 + hx}$ – fleablood Mar 21 '24 at 05:43

1 Answers1

1

You're so close! You just need to find a common denominator for the fractions.

$$12x+6h-\frac{6}{h(x+h)}+\frac{6}{xh}$$ $$= 12x+6h+\frac{-6x + 6(x + h)}{hx(x+h)}$$ $$= 12x+6h+\frac{6h}{hx(x+h)}$$ $$= 12x+6h+\frac{6}{x(x+h)}$$

Now, take the limit as $h \to 0$.

Dan
  • 14,978