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C4 differentation

Hi, i get that you have to use a^x ln a, but what do i do with the 200? because in differentiation the constant term is 0? i don't get it? even if you use the product rule and take u=200 and du/dx would be 0? pls help..

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@Zacken @Philip-flop @RDKGames

(edited 6 years ago)

Reply 1

RDKGames

Study Forum Helper

Original post by Jane122

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@Zacken @Philip-flop @RDKGames

0.9^t=e^{t\ln(0.9)}

Reply 2

Jane122

Original post by RDKGames

0.9^t=e^{t\ln(0.9)}

i don't understand where you got that from? why would it be e^ ? also, what happens to the 200?

Reply 3

RDKGames

Study Forum Helper

Original post by Jane122

i don't understand where you got that from? why would it be e^ ? also, what happens to the 200?

Because e differentiates to itself, so its the ideal base for the logarithm. Use some basic log rules and exponent facts to ensure the results are equal.

The 200 obviously stays there. Just differentiate

200e^{t\ln(0.9)}

(edited 6 years ago)

Reply 4

Jane122

Original post by RDKGames

Because e differentiates to itself, so its the ideal base for the logarithm.

The 200 obviously stays there. Just differentiate

200e^{t\ln(0.9)}

i seriously do not understand why the 200 stays there?? :/ also, yes but the rule is y=a^x then dy/dx= a^x ln a

where on earth is e^x in it?

Reply 5

RDKGames

Study Forum Helper

Original post by Jane122

i seriously do not understand why the 200 stays there?? :/ also, yes but the rule is y=a^x then dy/dx= a^x ln a

where on earth is e^x in it?

That rule is an alternative. Clearly it has ln in it so e mustve been used in deriving it. Anyway, I dont have much free time right now so I'll just say that using your rule there, consider the fact that 2e^x differeniates to 2e^x. Does the 2 disappear? No it does not. Same principle for your question.

Generally, y=af(x) means y'=af'(x) from basic rules you encounter at C1 so clearly your 200 doesnt disappear

(edited 6 years ago)

Reply 6

Jane122

Original post by RDKGames

i think i understand, its not 100% clear hm
thank you

Reply 7

davros

Original post by Jane122

i think i understand, its not 100% clear hm
thank you

I think you have 2 issues here:

1) if a constant 'a' multiplies a function, then the derivative is just the constant times the derivative of that function. This makes sense because the derivative just measures the gradient of (the tangent to) the curve at a point, so if all your y values are multiplied by 'a' then so is the derivative.

You can actually see this from the product rule as you suggested - it's just unnecessarily long-winded - if y = au where a is a constant, then

dy/dx = a (du/dx) + u(da/dx)

But the derivative of a constant is 0 i.e. da/dx = 0, so you're just left with

dy/dx = a (du/dx)

2) If you want to differentiate a^x then you can just use the rule from your formula book that says its derivative is (a^x)ln a.

However, you can also say that if

y = a^x

then

ln y = ln(a^x) = x(ln a)

by one of the standard laws of logs. Now using the fact that e^x and ln x are inverse functions you can write:

ln y = x (ln a) \implies y = e^{x (ln a)}

so you have an alternative expression for y. Since ln a is just a constant, you can use your standard rule for differentiating

e^{kx}

to differentiate y, which gives you the same answer as your formula book rule.

does this help?

Reply 8

Philip-flop

Original post by Jane122

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Zacken Philip-flop @RDKGames

Hi Jane, these types of differentiation had me confused at first too. I think it's cos the Edexcel endorsed textbook doesn't cover it too well. Luckily, RDKGames is one of the people who managed to teach me - I seriously can't thank him enough, he is actually some sort of Maths wizard who knows instantly where everything has derived from.

Anyway.. as you already know, to differentiate an a^x term we use the general formula