# Thread: Question on derivative question involving quotient & chain rule

1. ## Question on derivative question involving quotient & chain rule

There are 2 questions involving derivatives that I have trouble getting to the answer for. The first one is $y = [(6-5x) / (x^2 - 1)]^2$

and the 2nd one is

$y = 36 / (3-x)^2$

$[2(6 - 5x)(5x^2 - 12x + 5)] / (x^2 - 1)^3$
&
$y = (8/3)x + 4$

It would be very much appreciated if anyone could provide step-by-step solutions to both problems.

2. Just in case a picture helps...

... where

... is the chain rule, and...

... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).

So what you've got here is two chains inside a product-rule (the legs-uncrossed version). That's just my preference - you could do it as a product (or even a quotient - yeughh!) inside a chain.

Anyway, if you feel up to filling in the blanks and simplifying, notice I've tweaked the third balloon along for the sake of the common denominator.

Hope that helps, or doesn't further confuse.

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3. Do you have a solution for the 2nd problem by any chance?

4. Hate to say this (after last time) but are you QUITE sure that's the right derivative?

5. For the 2nd question?

6. Oh yes

7. Oh damn, the answer must have confused you because it's not the answer to the derivative directly, but an equation where the point is (0,4).

Because the question is to find an equation of the tangent line to the graph of the function at the given point. I'm terribly sorry for overlooking that!

8. Oh, fine. The derivative is the first step, it's not a hard one, just the chain rule in a simple way. I'll post a pic just in case it helps, but go ahead and try it, then I guess you know to plug in zero as the x-value in the derivative function, to find the slope of the tangent line...

No need to simplify or anything, just plug zero (as the value of x) into the derivative and you have the slope of the line. Now, the y-value where x is zero has a special role in one form of the line's equation, no?

9. Oh yeah, the second part is easy. It's only the derivative part that I'm concerned about.

Could you please include all the steps for the first question? I don't know how they got to that long answer, mine doesn't match us.

10. Well, can you see that the denominators match?

11. Put it another way... can you see why the final (correct) denominator is (x^2 - 1)^3
?

12. My denominator matches that of the answer, but I must've messed up somewhere on the calculations of the numerator because they don't match.

13. OK, you're ahead of me, so I don't know for sure if the numerator will match*, but do you get (for the numerator) the following...

$2(6 - 5x)(-5)(x^2 - 1) + (6 - 5x)^2 (-2) (2x)$

If not, read off from the bottom row of balloons

*Yep, it does

14. Originally Posted by tom@ballooncalculus
OK, you're ahead of me, so I don't know for sure if the numerator will match, but do get (for the numerator) the following...

$2(6 - 5x)(-5)(x^2 - 1) + (6 - 5x)^2 (-2) (2x)$

If not, read off from the bottom row of balloons
I got all of the above values except the $(x^2 - 1) + (6 - 5x)^2$. How did you get these two?

15. Well, I can only suggest checking yours against the bottom row of balloons and checking that the balloons are following the product-rule, as they should. The product rule may be hard to see amongst all the chain rule details so I might simplify...

By the way, beware the 'tweak', third balloon along, which is for the sake of the common denominator.

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