Question

Find any stationary points in the interval 0
11 years ago

Answer 1

Please also explain to me how how you use the chain rule with this.

Answer 2

\( \Large
y = (\sin(x))^2 + 2\cos(x) \\ \Large
\frac {dy}{dx} = \frac{d}{dx}(\sin(x))^2 + \frac {d}{dx}(2*\cos(x)) \\ \Large
~~~~ = 2(\sin(x))^1 * \frac {d}{dx}(\sin(x)) - 2\frac {d}{dx}\cos(x) \\ \Large
~~~~ = 2\sin(x)\cos(x) - 2\sin(x) \\ \Large
~~~~ = 2\sin(x)(\cos(x) - 1) \\
\)
To find stationary points, set \(\Large \frac {dy}{dx} = 0 \) and solve for x.

Answer 3

So one of the sin x's is treated differently from the other?

Answer 4

Not sure of the question. But to differentiate (sin(x))^2 we first apply the power rule which says the derivative of (something)^n = n*(something^(n-1) and then we apply the chain rule by multiplying the above by derivative of (something).
So d/dx (sin x)^2 = 2(sin x)^1 * d/dx(sin x) = 2sin(x)cos(x)

Answer 5

I understand, thank you very much!

Answer 6

If it would help you can make use of substitution:
If y = (sin x)^2 find dy/dx.
Let u = sin x
y = u^2
dy/dx = dy/du * du/dx
= d/du(u^2) * d/dx(sin x)
= 2u^1 * cos(x)
= 2sin(x)cos(x) (by putting u back as sin x)

You are welcome.

Answer 7

That's the first someone tried teaching me the substitution thing that I actually understood! Wow thanks and thanks again!

Answer 8

Glad to be able to help. And you are very welcome. :)