How do i use the chain rule twice when differentiating?

Ok so for this question we'll use this example: y = (e-2x^2 + 2)1/5
We start off by making the expression inside the brackets equal to u. In other words u = e-2x^2 + 2. This also means y = u1/5
Using the chain rule, we need to find du/dx and dy/du. dy/du = 1/5u-4/5. Now to find du/dx we must use the chain rule again as e-2x^2 has a power within a power.
So lets say z = e-2x^2 . You can use any letter but for this example i'll use z. And v = -2x2. Using the chain rule we get: dz/dx = -4xe-2x^2
Now we can go back to when we used the chain rule the first time to differentiate the whole expression. du/dx = -4xe-2x^2 and dy/du = 1/5u-4/5. Multiplying these together and replacing u with e-2x^2 + 2 gives us our final answer. dy/dx = 1/5(e-2x^2 + 2)-4/5-4xe-2x^2
Simplifying gives us dy/dx = -4xe-2x^2/5(e-2x^2 + 2)4/5

Answered by Indiya L. Maths tutor

4094 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

You are given the equation y=x^2. Determine whether or not the equation has any maximums or minimums and identify them (whether they are maximums or minimums).


A curve has the equation y=3x^3 - 7x^2+52. Find the area under the curve between x=2 and the y-axis.


2 log(x + a) = log(16a^6) where a is a positive constant. How do I find x in terms of a?


The point P lies on a curve with equation: x=(4y-sin2y)^2. (i) Given P has coordinates (x, pi/2) find x. (ii) The tangent to the curve at P cuts the y-axis at the point A. Use calculus to find the coordinates of the point A.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences