logarithms and derivatives-

[Nomvalt]

Differentiate the following:









But I don't understand how you get from:



to:



Please somebody show complete correct working out to help me understand. Where did the 3 come from?
Excuse my ignorance.

[nala]

The 3 appears as this is an application of the chain rule. You must multiply the differentiated function by the derivative of the expression in the bracket, which in this case is 3.

For example,

y= 2(2x-1)^3
dy/dx= 2(3)(2x-1)^2 * (2)
        =12(2x-1)^2

I hope that makes sense! (sorry, I don't know how to use latex)

[Nomvalt]

The 3 appears as this is an application of the chain rule. You must multiply the differentiated function by the derivative of the expression in the bracket, which in this case is 3.

For example,

y= 2(2x-1)^3
dy/dx= 2(3)(2x-1)^2 * (2)
        =12(2x-1)^2

I hope that makes sense! (sorry, I don't know how to use latex)

I think that I get what your trying to say. Though, I'm not sure how I've managed to get the correct answer for most of the questions in the exercise that I'm doing without keeping that close to mind.

[nala]

Hmm, if there is no coefficient of x inside the brackets (eg. (x+1) (x-5)), then the derivative is 1, therefore it stays the same. Perhaps the earlier questions all have expressions similar to these.

[GerrySly]

You just apply the chain rule *again*

Let

Then you are after

So using the chain rule you have

Now and which gives us

[Nomvalt]

Hmm, if there is no coefficient of x inside the brackets (eg. (x+1) (x-5)), then the derivative is 1, therefore it stays the same. Perhaps the earlier questions all have expressions similar to these.

The above is an example of one of the questions I did before that I got right despite incorrect working out. I got the right answer despite not showing the chain rule to find Notice that is wrong, however, the final answer obtained is correct ().

Notice that I also forgot to substract the exponent by 1 here:


I completely forgot the -2 exponent also and acted as though I was just moving "-2" as an exponent to the front of "log_e(5x + "

Another example is:









In this I forgot  all about subtracting 1 from the exponent of 4 in:

and like in my other example, I removed the original exponent of 4 altogether as if I was merely just moving to the front of


In both cases the values I got for were incorrect, however it did not affect the final correct result for

Could someone explain all this? I feel like I've discovered something new...

[GerrySly]

But you did apply it correctly? You don't need to subtract one from the indice if it's a log, doing what you did is perfectly fine.

But if you look at your original one there is a fraction inside the brackets so it's...



Just remember the rule, If then . If there's a co-efficient out the front of the log or the log is to a power, take it to the front, factorise it out then perform the rule and multiple the remaining fraction by the co-efficient. i.e. then it'll be

[Nomvalt]

But you did apply it correctly? You don't need to subtract one from the indice if it's a log, doing what you did is perfectly fine.

But if you look at your original one there is a fraction inside the brackets so it's...



Just remember the rule, If then . If there's a co-efficient out the front of the log or the log is to a power, take it to the front, factorise it out then perform the rule and multiple the remaining fraction by the co-efficient. i.e. then it'll be

  ....hmmm, I guess that makes sense.

[Ilovemathsmeth]

The 3 comes from the chain rule. Remember that when you differentiate using the chain rule, you bring the power to the front, multiplying with the constant and subtract one from the old power to give the new power. You also multiply by the derivative of the inside function.

I think you're confused because you wanted to use the chain rule formally to differentiate that part of the function. You could have done this with the use of 'w' and 'h' or something. I always found it more confusing however, to keep using 'w' and 'h' etc so I used the shorter chain rule method - explained in the book - which isn't really shorter, it just omits the use of 'let u = this' etc. This helps you do these questions faster.