how do i work out the value of dy/dx at P when it is 3, using..?

Question

"the curve C has the equation y=4x² + (5-x)÷x, x≠0. The point P on C has x-coordinate 1. Show that the value of dy/dx at P is 3"

Answer 1

First you'll need to find dy/dx, which is the first derivative of curve C.

y=4x² + (5-x)/x

dy/dx = 8x -5/x²

Now we need to find out what dy/dx equals at point P. The x-coordinate of P is 1, so we plug in 1 for x and solve:

dy/dx= 8(1) -5/(1)² = 8 - 5 = 3

Answer 2

y=4x² + (5-x)/x
f(x) = 5-x, g(x) = x
[ f'(x)g(x) - f(x)g'(x) ] / g(x)^2
y' = 8x + [ f'(x)g(x) - f(x)g'(x) ] / g(x)^2
y' = 8x + ((-1)x - (5-x)*1) / x^2
y' = 8x + [ (-x + x - 5) / x^2 ]
y' = 8x - 5/x^2
y' = 8*1 - 5/1^2
y' = 8 - 5/1
y' = 8 - 5
y' = 3
dy/dx = 3