@Jessycar1BrumBeep From the teaching point of view, it is important to understand where the formula dy/dx = (f '(x)) / (f(x)) comes from. I often find a lot of people do not know this and this was the point of the exercise. Thanks for your comment though.
I, (we) use the standard results when we differentiate a single term, such as ln x, ln t, ln p, ln q etc. So the chain rule is used when we have ln (several terms). Making y=ln(t) allows us to find dy/dt and then this is used to find dy/dx via the chain rule.
+ExtremeTG To get A* you need 90%+ in Core 3 and 4. I think I should of done more work in applying my knowledge rather than learning facts. Definitely dont just do AQA past papers, but also try Edexcel and Ocr
The standard results are used. Differential of ln x with respect to x is 1/x. The differencial of ln z with respect to z would be 1/z, the differential of ln t with respect to t would be 1/t and so on.
Thank you! Need to rewrite my notes now but I've finished C3... that's 1 out of 7 exams I'm sitting this Jan I've covered the unit entirely for. And 1 out of 7 I can start past papers on :) Owe it to you!
This is so awesome..Thanks. I've been trying to understand this for days but this video simplified all that in minutes.....ha.haa N the acsent too is clear which improves concentration
@MrValzen when you have a log function to the power of something you take the power and multiply it by the log. e.g. y=ln(x^2-1)^1/2=1/2ln(x^2-1) dy/dx=x/(x^2-1)
x=3/2 i think Use log rules to bring the 3 into the lnx giving e^ln x^3=27/3 then since e to the power of the natural log of anything equals the thing inside the natural log (in this case the x^3) you'd get x^3=27/8 then cube root to get x=3/2
multiplying dy/dt by dt/dx does not mean you do that. dy/dt is 1/(3x^2-1) and dt/dx is the differential of (3x^2-1), as we replaced the (3x^2-1) by t. Therefore it should be 1/(3x^2-1) multiplied by 6x. Hope that made sense.
