How do I differentiate and integrate powers of x?

For differentiation, we can follow a simple method for differentiating y with respect to x. It may be the case that you have special numbers such as ex you need to differentiate and integrate but these formulas are easily findable online.
The general rule of differentiation is that for y = axb, dy/dx = abxb-1 , so for example if we had y = 3x2 then dy/dx = 6x. We have multiplied the 3 by 2 and subtracted the original 2 in order to complete the differentiation.
For integration, you would do the reverse, so that for dy/dx = axb you would integrate it to get y = (axb+1)/b+1 + c, and we can easily check this. But we need to ask, where did that constant come from? This is easy, as for any constant c, c = c * 1 = cx0 as any number to the power of 0 is 1! So when the 0 carries over to the left hand side it removes the whole number. So just to check, by differentiating the c we remove it, we times our fraction by b+1 removing the demominator, and then b+1-1 = b which gets us back to axb! So for example, if we had dy/dx = 3x2+ 2x + 2, we would end up with y = (3x3)/3 + (2x2)/2 + (2x1)/1 + c leading to y = x3 + x2 + 2x + c! If given x and y values where this holds we can quickly work out c. If we wanted to find the area between two x values on a curve we would substitute them in here.

TT
Answered by Terry T. Maths tutor

2764 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the x-values of the turning points on the graph, y=(3-x)(x^2-2)


Points P and Q are situated at coordinates (5,2) and (-7,8) respectively. Find a) The coordinates of the midpoint M of the line PQ [2 marks] b) The equation of the normal of the line PQ passing through the midpoint M [3 marks]


Find the equation of the tangent at x=1 for the curve y=(4x^2+1)^3


How would I use implicit differentiation to differentiate functions such as: y=tan^-1(ax^2+b) in the form of dy/dx=.....?


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