Lectures

Differentiation: Exercise 1 – Calculus Methods

by Batool Akmal
(1)

Questions about the lecture
My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Differentiation Exercise Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake
    Transcript

    00:01 Welcome back.

    00:02 I've hope you've enjoyed having a go on the questions and the exercises.

    00:06 We're now just going to go through it together to see whether you've done it correct or incorrectly.

    00:11 And also to see if there is any problems that you've had so that we can help you out with them.

    00:17 So looking at our first question.

    00:19 Nice and straight forward.

    00:21 Find the general gradient of the curve.

    00:24 Remember what we said about general gradient, it's just the gradient.

    00:27 It's just the gradient as a function.

    00:29 You don't need to substitute any numbers into it.

    00:32 You don't need to find any tangents or normals.

    00:34 You just have to find the gradient as an expression.

    00:37 So let's have a look at this equation we've got y = root x + 2x - 1.

    00:42 Now recall what we said if we had any root x's or any powers, any indices that are down as denominators, we can bring them up using rule of indices.

    00:53 But in this case all we have to do is change the square root.

    00:56 So we can rewrite this as y = x to the half.

    01:02 Plus 2x minus 1.

    01:06 And the general gradient dy by dx.

    01:08 Bring the power down.

    01:12 Decrease the power by 1.

    01:14 Same again.

    01:16 Bring the power down.

    01:18 Which is just 1, 2x, 1 - 1 and the constant just disappears.

    01:24 So the answer to this will be the half x to the minus the half plus 2.

    01:31 Because x to the 0 is just going to be 1.

    01:34 And again you can take that down.

    01:37 So 2x to the half plus 2.

    01:40 Or you could rewrite this as a root to see we write this as 1 over 2 root x plus 2.

    01:47 And again this is your general gradient, dy by dx.

    01:50 Which means to find a particular gradient you just have to replace this x with different values or different points in the curve that you are looking at.

    02:00 And you will find your gradient as a numerical value.


    About the Lecture

    The lecture Differentiation: Exercise 1 – Calculus Methods by Batool Akmal is from the course Calculus Methods: Differentiation.


    Included Quiz Questions

    1. y=x^(-1/2)
    2. y=x^(1/2)
    3. y=x^(-1)
    4. y=x^(-2)
    5. y=x^(2)
    1. dy/dx=(-1/2)x^(-3/2)
    2. dy/dx=(-1/2)x^(1/2)
    3. dy/dx=(-1/2)x^(-1/2)
    4. dy/dx=(1/2)x^(-3/2)
    5. dy/dx=(-1/2)x^(3/2)

    Author of lecture Differentiation: Exercise 1 – Calculus Methods

     Batool Akmal

    Batool Akmal


    Customer reviews

    (1)
    5,0 of 5 stars
    5 Stars
    5
    4 Stars
    0
    3 Stars
    0
    2 Stars
    0
    1  Star
    0