Find d y d x for the following: y = - 10 x 12 - 8 x 10

Given, y = - 10 x 12 - 8 x 10 We have to find d y d x for the given equation. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables. We know that, If y = a x n , then d y d x = n × a × x n - 1 Now, y = - 10 x 12 - 8 x 10 and d y d x = 12 × - 10 × x 12 - 1 - 8 × 10 × x 10 - 1 = - 120 x 11 - 80 x 9 [Differentiate each term separately] Hence, d y d x = - 120 x 11 - 80 x 9

Find d y d x for the following: y = 4 x 2 - 12 x 3 + 5 x 4

Given, y = 4 x 2 - 12 x 3 + 5 x 4 We have to find d y d x for the given equation. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables. We know that, If y = a x n , then d y d x = n × a × x n - 1 Now, y = 4 x 2 - 12 x 3 + 5 x 4 and d y d x = 2 × 4 × x 2 - 1 - 3 × 12 × x 3 - 1 + 4 × 5 × x 4 - 1 = 8 x - 36 x 2 + 20 x 3 [Differentiate each term separately] Hence, d y d x = 8 x - 36 x 2 + 20 x 3

Find d y d x for the following: y = - 8 11 x 4 + 2 7 x 3 - 3 4 x 2

Given, y = - 8 11 x 4 + 2 7 x 3 - 3 4 x 2 We have to find d y d x for the given equation. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables. We know that, If y = a x n , then d y d x = n × a × x n - 1 Now, y = - 8 11 x 4 + 2 7 x 3 - 3 4 x 2 and d y d x = - 4 × 8 11 × x 4 - 1 + 3 × 2 7 × x 3 - 1 - 3 4 × 2 × x 2 - 1 = - 32 11 x 3 + 6 7 x 2 - 3 2 x [Differentiate each term separately] Hence, d y d x = - 32 11 x 3 + 6 7 x 2 - 3 2 x

Find the co-ordinates of the points at which the curve with equation y = 1 4 x 4 - 3 2 x 2 has gradient zero.

We have to find the co-ordinates of the points at which the gradient of the curve with equation y = 1 4 x 4 - 3 2 x 2 has gradient zero. The derived function gives the gradient at the point where the x co-ordinate is x . We know that, If y = a x n , then d y d x = n × a × x n - 1 Now, y = 1 4 x 4 - 3 2 x 2 and d y d x = 1 4 × 4 × x 4 - 1 - 3 2 × 2 × x 2 - 1 = x 3 - 3 x Given that, gradient d y d x = 0 ⇒ x 3 - 3 x = 0 ⇒ x x 2 - 3 = 0 ⇒ x = 0 , x 2 - 3 = 0 ⇒ x = 0 , x = 3 Substituting x = 0 , in the equation of the curve, we get, y = 1 4 0 4 - 3 2 0 2 = 0 Substituting x = 3 , in the equation of the curve, we get, y = 1 4 3 4 - 3 2 3 2 = 9 4 - 9 2 = - 9 4 Hence, the co-ordinates of the point are 0 , 0 and 3 , - 9 4 .

E M B I B E

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Algebra>Chapter 18 - Curved Graphs>Exercise 18.12

Karen Morrison and Nick Hamshaw Solutions for Exercise 12: Exercise 18.12

Author:Karen Morrison & Nick Hamshaw

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 12: Exercise 18.12

Attempt the free practice questions from Exercise 12: Exercise 18.12 with hints and solutions to strengthen your understanding. Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition solutions are prepared by Experienced Embibe Experts.