How to differentiate using the chain rule

The chain rule is a rule of differentiation that states: (f(g(x))'=f'(g(x))g'(x)

In effect, the chain rule allows us to split up a complicated equation into smaller more manageable chunks. Rather than differentiating f(g(x)) all in one go, all we need to do is differentiate f(.) - giving us f'(g(x)) -  and then differentiate g(.) - giving us g'(x). We can then multiply these together to find the derrivative of the original equation.

Consider the following example:

e3x+8=y

At first this equation may seem complicated and difficult to differentiate, in fact, it's quite easy. All we need to do is apply the chain rule.

In this case, we can see that the equation is made up of two "chunks" which we can seperate out.

Firstly there is the simple linear equation:

g(x)=3x+8

and then there is also the exponential function:

eg(x)=f(g(x))

Then it is easy to see that f(g(x))=y (just substitute g(x)=3x+8 into the equation eg(x))  This is an equation of exactly the form we need in order to use the chain rule.

Now we have a much simpler problem, we just need to differentiate each "chunk"

g'(x)= 3     (this is just a linear equation so this easy to check)

f'(g(x))=eg(x)   (Since the derrivative of the exponential function is simply the very same function).

We can then multiply these together to find the derrivative of the original equation:

dy/dx = 3e3x+8

AT
Answered by Anthony T. Maths tutor

3443 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

In a science experiment a substance is decaying exponentially. Its mass, M grams, at time t minutes is given by M=300e^(-0.05t). Find the time taken for the mass to decrease to half of its original value.


Differentiate x^2+4x+9.


(1.) f(x)=x^3+3x^2-2x+15. (a.) find the differential of f(x) (b.) hence find the gradient of f(x) when x=6 (c.) is f(x) increasing or decreasing at this point?


Express 1/(1+2x)(1-x) in partial fractions


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences