1. ## Differentiation Stuff

x=y^2-3ln(2y)

a) show that dy/dx = y/(2y^2-3)

b) find equation for tangent to curve where y=1/2
give answer in form ax+by+c=0 where a, b, c are all integers

Have done a) cannot do b) can someone help me please?

Should have said, can do b), but my answer is never an integer, can someone see wot i am doing wrong. i get x=1/4 and dy/dx = -1/5 are these wrong?

2. Originally Posted by Kiwi
x=y^2-3ln(2y)

a) show that dy/dx = y/(2y^2-3)

b) find equation for tangent to curve where y=1/2
give answer in form ax+by+c=0 where a, b, c are all integers

Have done a) cannot do b) can someone help me please?
To get a point on the line:

Substitute y = 1/2 into $\frac{dy}{dx} = \frac{y}{2y^2 - 3}$ to get the gradient of the tangent.

Substitute y = 1/2 into $x = y^2 - 3 \ln (2y)$ and solve for x. You now have the coordinates of a point on the tangent.

So now you have a gradient and a point. Can you use these two things to get the equation of the tangent line?