Question

Find the Derivative
r=sin(root2theta)

Answer 1

Well, two layers here.
\[\sin(---)\]and
\[\sqrt{2\theta} \]So we just need to take the derivative of each one and then multiply the results.

Answer 2

\[\frac{ 1 }{ 2 }(\sin 2x)^{-\frac{ 1 }{ 2 }}(2\cos 2x)\]

Answer 3

Oh, is sin also square rooted?

Answer 4

No

Answer 5

Just the 2x, right?

Answer 6

Yes

Answer 7

Alright then yeah, we need to try again. So the first layer is just sin(----), which becomes cos(----). Our sin disappears, we dont have a sin factor anywhere left in this problem. So we have cos(---) multiplied by the derivative of sqrt(2x).So now all we need is the derivative of sqrt(2x) and we'll multiply it by cos(sqrt2x)

Answer 8

\[(\cos \sqrt{2x})(\frac{ 1 }{ 2\sqrt{2x} })\]

Answer 9

Forgot an extra 2. You brought the power down as a multiplication and lowered the power by 1, but you have to multiply it by the derivative of 2x that is inside of the root. So you should have an extra multiplication of 2. Do you see why?

Answer 10

Yes, so its \[\frac{ 2 }{ 2\sqrt{2x} }\]?

Answer 11

Right, so the 2's cancel out.

Answer 12

So the complete answer is \[\frac{ \cos \sqrt{2x} }{ \sqrt{2x} }\]

Answer 13

Yes : )

Answer 14

Thank you

Answer 15

\[r=\sin(\sqrt {2\theta}) \rightarrow \frac{dr}{d \theta} = \cos \sqrt{2 \theta} \frac{d}{d \theta} (2 \theta)^{ \frac{1}{2}}\]
\[= \cos \sqrt{2 \theta} \frac{d}{d \theta} \frac{1}{2} (2 \theta)^{ \frac{-1}{2}} \times 2\]
\[= \cos \sqrt{2 \theta} \frac{d}{d \theta} \ (2 \theta)^{ \frac{-1}{2}} \]
\[= \cos \sqrt{2 \theta} \ \frac{1}{ (2 \theta)^{ \frac{1}{2}} }\]
\[= \cos \sqrt{2 \theta} \ \frac{1}{ \sqrt{(2 \theta)} }\]
\[= \frac {\cos \sqrt{2 \theta}}{\sqrt{2 \theta}} \]