1. ## Differentiation.

1) The point at which the slope of the curve y = 2x2 −1 is equal to 2 , is

2x^2 = 1+2
x^2=3/2
x= -1/2 , or x = 1/2

3). The derivative of y = cos2θ is

− 2cosθ sinθ

Moderator edit: Refer forum rule #8: http://www.mathhelpforum.com/math-he...ng-151418.html

2. Originally Posted by r-soy
1) The point at which the slope of the curve y = 2x2 −1 is equal to 2 , is

2x^2 = 1+2
x^2=3/2
x= -1/2 , or x = 1/2
I'm not sure to which question these particular calculations belong

The slope of the curve is calculated by the first derivative:

$\displaystyle f(x)=2x^2-1~\implies~f'(x)=\frac{dy}{dx}=4x$

According to he question you are asked to determine the value of x if $\displaystyle 4x=2$

...

3). The derivative of y = cos2θ is

− 2cosθ sinθ

...
You forgot to use the chainrule:

$\displaystyle y = \cos(2\theta)~\implies~\frac{dy}{d \theta} = -\sin(2\theta}) \bold{\cdot 2}$

With $\displaystyle \sin(2\theta) = 2 \cdot \sin(\theta) \cdot \cos(\theta)$ you'll get:

$\displaystyle \frac{dy}{dx}=-4 \cdot \sin(\theta) \cdot \cos(\theta)$