# Thread: How do you work out the exact gradient when y = 3cos2x?

1. ## How do you work out the exact gradient when y = 3cos2x?

Hi, I Know what to do, find the first derivative. But i get stuck on a line which can't find the exact value.

Question: Find the gradient of the tangent to the curve y = 3cos2x at the point where x = pi/6.

dy/dx = -6sin2x

y'(pi/6) = -6sin(2*pi/6)

=-6sin(2pi/6)

how do I simplify that? I'm totally lost as of what to do next.

PS: On another note what does d/dx mean as compared to dy/dx?

Thanks,

MBB

2. Originally Posted by MathsBeforeBedtime
Hi, I Know what to do, find the first derivative. But i get stuck on a line which can't find the exact value.

Question: Find the gradient of the tangent to the curve y = 3cos2x at the point where x = pi/6.

dy/dx = -6sin2x

y'(pi/6) = -6sin(2*pi/6)

=-6sin(2pi/6)

how do I simplify that? I'm totally lost as of what to do next.

PS: On another note what does d/dx mean as compared to dy/dx?

Thanks,

MBB
You may need to review your unit circle to find $\displaystyle \sin(\pi/3)$...

Also, $\displaystyle \frac{\,d}{\,dx}\left[\cdots\right]$ is a way of saying that you're differentiating something. So for example $\displaystyle \frac{\,d}{\,dx}\left[x^2+2x+1\right]$ means that you should differentiate $\displaystyle x^2+2x+1$ with respect to $\displaystyle x$.

3. Originally Posted by Chris L T521
You may need to review your unit circle to find $\displaystyle \sin(\pi/3)$...

Also, $\displaystyle \frac{\,d}{\,dx}\left[\cdots\right]$ is a way of saying that you're differentiating something. So for example $\displaystyle \frac{\,d}{\,dx}\left[x^2+2x+1\right]$ means that you should differentiate $\displaystyle x^2+2x+1$ with respect to $\displaystyle x$.
why pi/3

4. Because $\displaystyle \frac{2}{6}= \frac{1}{3}$. If you couldn't see that, no wonder you have problems with this!

Also $\displaystyle \frac{\pi}{3}$ radians is the same as $\displaystyle \frac{180}{3}= 60$ degrees. And 60 degrees is the measure of each of the angles in an equilateral triangle. Imagine an equilateral triangle with each side of length 1. Drop a perpendicular from one vertex to the opposite side. That will also bisect the opposite side so you now have two right triangles with one leg of length 1/2 and hypotenuse of length 1. By the Pythagorean theorem, $\displaystyle 1^2= (1/2)^2+ b^2$ where b is the length of the other leg- the perpendicular. Then $\displaystyle b^2= 1- 1/4= 3/4$ so $\displaystyle b= \sqrt{3}//2$. Sine is "opposite side over hypotenuse so $\displaystyle sin(\pi/3)= sin(60)= \sqrt{3}/2$.