1) y= (2x+5)(x-1)
First we expand the brackets, and simplify :
2x^2 - 2x + 5x -5
y = 2x^2 + 3x -5, now we can differentiate so :
dy/dx = 4x + 3
I will show the 2nd example, in my set of image notes.
Application of Differentiation to Equations of Tangents
If we have a curve, and a tangent to some point on the curve. We can find the equation of the tangent. How ?
If we know the coordinates of that point. Firstly we differentiate the curve, and substitute the x-coordinate to get the gradient of the curve at the point. Say, if the tangent is perpendicular to the curve. We know that the gradient of this tangent, will be -1 divided by the gradient of the curve, we just got. Using the gradient, and the coordinates, we can use the y -y1 = m(x-x1) formula to find out the equation.
This question is taken from the Edexcel C1 Jan 2006 Paper. (Question 9)
|Sorry for the bad quality|
|I have ignored part a).. the rest of the question is more relevant to the post.|
|Notes on Differentiation with Four Examples|
|Application of Differentiation for Tangents,Gradients and Normals|