# Thread: Trigonometric derivatives?

1. ## Trigonometric derivatives?

y=5x csc x Did I differentiate this correctly

5x-cscxcotx+cscx 5 I used the product rule

My second question is

y=-cscx-sin(x)

2. ## Re: Trigonometric derivatives?

Originally Posted by homeylova223
y=5x csc x Did I differentiate this correctly

5x-cscxcotx+cscx 5 I used the product rule
There should only be two terms when you use the product rule - not three. I suspect you mean $-5x \csc(x)\cot(x) + 5\csc(x)$ which is the correct.

Spoiler:
My second question is

y=x^2 sin(x)+2x cos(x) This is what I did

(x^2)cos(x)+2x -sin(x)+2x sin(x)+2 cos(x) Have I done this correctly
You need to pay more attention to your signs and how you show them: $a \times -b$ must never be written as $a-b$ but instead $-ab$

Spoiler:
$x^2 \cos(x) + 2x\sin(x) - 2x \sin(x) + 2\cos(x) = \cos(x)(x^2+2)$

edit: I put the original second question in a spoiler since the OP edited it out

For your (new) second equation you can take the derivatives separately. $y = -\csc(x) - \sin(x)$

If you don't know the derivative of -csc(x) then you can write it as $-(\sin(x))^{-1}$ and use the chain rule.

3. ## Re: Trigonometric derivatives?

Ah I have not learned the chain rule yet only the product,quotient, and power rule.

But this is what I did

-csc(x)-cos(x)+cot(x)csc(x)-sin(x) I tried using the product rule

4. ## Re: Trigonometric derivatives?

Is your question $y = -\csc(x) \times - \sin(x)$ or $y = -\csc(x) - \sin(x)$ -- the latter is subtraction.

Since you've used the product rule I suspect it's the former. If so use the fact that $\sin(x)\csc(x) = 1$

$\csc(x)\cos(x)-\cot(x)\csc(x)\sin(x)$
That can be simplified:

$\csc(x)\cos(x) - \cot(x)\csc(x)\sin(x) = \dfrac{\cos(x)}{\sin(x)} - \dfrac{\cos(x)\sin(x)}{\sin(x)\sin(x)} = 0$

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-csc(x)-cos(x)+cot(x)csc(x)-sin(x) = $-\csc(x)-\cos(x)+\cot(x)\csc(x) - \sin(x)$ which clearly isn't right.