The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Differentiation and the chain rule

Hi,
I'm currently doing C4 revision, I have left it to the last minute and forgot nearly everything maths related -.-
a) please could someone go through how I would differentiate ln(x/2)
b) Also, please could someone briefly explain (or point me to a link that does) the chain rule (I want to know it because integration by recognition is based on it)

Thanks for any help in advance
Reply 1
Avatar for Dr Grip
Dr Grip
The chain rule involves differentiating an inside function and an outside function. So for example, your question:

Unparseable latex formula:

y= \ln \frac{x}{2}}



The inside function,
Unparseable latex formula:

u= \frac{x}{2}}


The outside function,
Unparseable latex formula:

y = \ln u}



Now the chain rule is:
dydx=dudx×dydu\frac{dy}{dx} = \frac{du}{dx} \times \frac{dy}{du} (i.e. imagine they are fractions)

dudx=12\frac{du}{dx} = \frac{1}{2}
dydu=1u=2x\frac{dy}{du} = \frac{1}{u} = \frac{2}{x}

dydx=12×2x=1x\frac{dy}{dx} = \frac{1}{2} \times \frac{2}{x} = \frac{1}{x}

You most likely won't need to list everything out like that after a few practicing.
Reply 2
Avatar for qgujxj39
qgujxj39
17
a) Use the identity ln(a/b) = ln(a) - ln(b), and then differentiate. You could also use the chain rule, and you'll get the same answer. Don't worry if your answer seems a bit weird or counter-intuitive - it's correct.

b) I'll describe the general method, then give a simple example.
Say you have a function of xx that you want to differentiate. Call this function yy. We want to find dydx\frac{\mathrm{d}y}{\mathrm{d}x}.
Suppose we can't find it straight away, but we see that we have a "function within a function" in our expression for yy. Call this uu. Then we have yy as a function of uu and uu as a function of xx. Then the chain rule tells us that
dydx=dydu×dudx\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}x}.
Note that once we've found dydu\frac{\mathrm{d}y}{\mathrm{d}u} as a function of uu, we'll want to re-express it as a function of xx.

An example: differentiate (x2+7)8(x^2 + 7)^8.
Let y=(x2+7)8y = (x^2 + 7)^8. It's clear that we have a "function within a function", in this case the bit inside the brackets. Let's call it u=x2+7u = x^2 + 7. Then we have y=u8y = u^8. We know how to differentiate both of these:
dydu=8u7=8(x2+7)7\frac{\mathrm{d}y}{\mathrm{d}u} = 8u^7 = 8(x^2 + 7)^7
dudx=2x\frac{\mathrm{d}u}{\mathrm{d}x} = 2x
and hence by the chain rule
dydx=dydu×dudx=16x(x2+7)7\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \times \frac{\mathrm{d}u}{\mathrm{d}x} = 16x(x^2 + 7)^7.
Reply 3
Avatar for kej817
kej817
OP
5
Thanks :biggrin: both explanations make sense, I understand now

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you