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Help with Differentiation

Hi,

Would somebody be able to explain why the derivative is Q1/Q2?

I know that the log rules means that differentiating lnQ2 means we get 1/Q2.

But, since we're treating Q1 as a constant, why doesn't it just disappear in the answer? Why does it get placed in 'Q1/Q2'?

Any help appreciated!

Diff.JPG
Q1 isn't a constant it's a coefficient. If it was + Q1 it would disappear. But much like 5x differentiates to 5, the coefficient doesn't get removed.

So when differentiating with respect to Q2, then it becomes Q1 multiplied by the differential which is 1/Q2 which you can simplify to Q1/Q2

Similarly if you differentiate 3sinx that will become 3cosx. The coefficient remains. But if it was sinx + 3 then in that case the 3 would disappear
ddbaln⁡b=ab′b\frac{d}{db} a\ln b = \frac{a b'}{b}
since the differential of b or Q2 (wrt q2) in this case is 1, it is just a/b or Q1/Q2
(edited 7 years ago)
Original post by AndrewSCO
Q1 isn't a constant it's a coefficient. If it was + Q1 it would disappear. But much like 5x differentiates to 5, the coefficient doesn't get removed.

So when differentiating with respect to Q2, then it becomes Q1 multiplied by the differential which is 1/Q2 which you can simplify to Q1/Q2

Similarly if you differentiate 3sinx that will become 3cosx. The coefficient remains. But if it was sinx + 3 then in that case the 3 would disappear


Thanks! Makes sense.

In that case, if the question was slightly different, e.g. 2Q1lnQ2, would the derivative be 2 multiplied by 1/Q2?
Original post by Exceptional
Thanks! Makes sense.

In that case, if the question was slightly different, e.g. 2Q1lnQ2, would the derivative be 2 multiplied by 1/Q2?


No (edit, misread), you may recall that (af(x))′=af′(x)(af(x))' = af'(x)(one of the first principles of differentiation that you're taught), so (2Q1lnQ2)′=2Q1(1Q2)(2Q_{1}lnQ_{2})' = 2Q_{1}(\frac{1}{Q_{2}}).
(edited 7 years ago)
Original post by Exceptional
Thanks! Makes sense.

In that case, if the question was slightly different, e.g. 2Q1lnQ2, would the derivative be 2 multiplied by 1/Q2?


close but it would be 2Q1/Q2, You are differentiating with respect to Q2 remember
ddbaln⁡b=ab′b\frac{d}{db} a\ln b = \frac{a b'}{b}

Try this question ddQ22Q1ln⁡4Q2\frac{d}{dQ2} 2Q1\ln 4Q2
Original post by Exceptional
Thanks! Makes sense.

In that case, if the question was slightly different, e.g. 2Q1lnQ2, would the derivative be 2 multiplied by 1/Q2?


Nope it would be 2Q1, because the coefficient is the whole thing (2Q1)lnQ2

Think of it as a coefficient and a function. So if you take 10x, x is the function and 10 is the coefficient. The derivative is the coefficient multiplied by the derivative for the function. So in this case it is 10 multiplied by derivative of x which is 1. Therefore the answer is 10.

IWith 2Q1lnQ2 (when deriving with respect to Q2), then (2Q1) is the coefficient and lnQ2 is the function. As you said, in this case Q1 is no longer a variable, it just acts as any random number, so 2 x Q1 is just another random number. It's like having (2 x 5)lnQ2, you don't only take the 2, you take the whole thing, just in that case you can simplify it further. So coef. multiplied by the derivative of the function is 2Q1 multiplied by the derivative which is 1/Q1.

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