Differentiate (with respect to x), y=2x^2+8x+5.

y= 2x^2 + 8x + 5 is a simple quadratic equation and can be easily differentiated. Consider it in different terms working from left to right: y converts to dy/dx. This just means ‘differentiate y with respect to x’ and is how we show we have differentiated the equation.
For the rest of the terms simply multiply the power that the x has by the number before the x and then minus 1 from the power. I.e. 2x^2 = (2x2)x^(2-1) = 4x^(1)=4x. Next term, 8x = 8x^1 = (8x1)x^(1-1) = 8x^0 = 8. Finally, 5 = 5x^0 = (5x0)x^(0-1) = 0. Notice how I have added powers that you would not normally show for explanation but you do not need to show this when writing in the exam. Also remember that x^0 is always 1.
Therefore replacing the original terms with the differentiated terms, we get:
dy/dx = 4x + 8 + 0 = 4x + 8

JD
Answered by Jack D. Maths tutor

1339 Views

See similar Maths Scottish Highers tutors

Related Maths Scottish Highers answers

All answers ▸

a) Factorise: 2x^2-72, and hence b) find the y-intercept of the line with the equation: y=(2x^2-72)/(4x-24)


Given that dy/dx = 6x*2 - 3x + 4 And y =14 when x=2. Express y in terms of x


y=x^3-3x^2+2x+5 a)Write down the coordinates of P the point where the curve crosses the x-axis. b)Determine the equation of the tangent to the curve at P. c)Find the coordinates of Q, the point where this tangent meets the curve again.


Determine for what values of c, f(x)=4x^2-(2c+8)x+4 has no real roots.


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning