Partial Derivative of 1/sin(y/2) with respect to x

[duo]

Homework Statement

d/dx 1/sin(y/2)

The Attempt at a Solution

this isn't an entire question, just looking for clarification about something.

i have been asked as part of a larger question to find the partial derivative of 1/sin(y) with respect to x. in this case you treat y as a constant, yes?

so d/dx of (sin(y/2) = cos(y/2)*dy/dx, since y is treated as a constant, dy/dx = 0, so d/dx of sin(y/2) = 0. if you use the quotient rule on the full equation, you divide by 0^2, which is 0, so you end up dividing by 0. Is this correct?

The full expression actually has

sin(x/2 + y/2)

above instead of 1. is it just a matter of rearranging this expression so that the bottom cancels out? or does d/dx sin(y/2) not actually evaluate to 0 in the first place?

thanks in advance!

 

[asd1249jf]

Homework Statement

d/dx 1/sin(y/2)

The Attempt at a Solution

this isn't an entire question, just looking for clarification about something.

i have been asked as part of a larger question to find the partial derivative of 1/sin(y) with respect to x. in this case you treat y as a constant, yes?

so d/dx of (sin(y/2) = cos(y/2)*dy/dx, since y is treated as a constant, dy/dx = 0, so d/dx of sin(y/2) = 0. if you use the quotient rule on the full equation, you divide by 0^2, which is 0, so you end up dividing by 0. Is this correct?

The full expression actually has

sin(x/2 + y/2)

above instead of 1. is it just a matter of rearranging this expression so that the bottom cancels out? or does d/dx sin(y/2) not actually evaluate to 0 in the first place?

thanks in advance!

I think you are making this way too complicated as it shouldn't be.

Well first of all, you mentioned about dividing by 0^2. Remember that you can't divide anything by 0.

If you were asked to take the partial derivative of

$$\sin(\frac{x}{2}+\frac{y}{2})$$

with respect to x, as you mentioned earlier, just treat y as a constant in this case, then take derivative.

For instance, derivative for $$\sin(\frac{x}{2}+\frac{1}{3})$$ is simply $$\frac{1}{2}cos(\frac{x}{2}+\frac{1}{3})$$

 

[duo]

right, but it's $$\frac{\sin({\frac{x}{2} + \frac{y}{2}})}{\sin({y/2})}$$

with respect to x. so you use the quotient rule to evaluate it.

but the expression on the bottom evaluates to 0 when you differentiate with respect to x? or am i totally wrong? also thank you for responding ;-)

 

[asd1249jf]

right, but it's $$\frac{\sin({\frac{x}{2} + \frac{y}{2}})}{\sin({y/2})}$$

with respect to x. so you use the quotient rule to evaluate it.

but the expression on the bottom evaluates to 0 when you differentiate with respect to x? or am i totally wrong? also thank you for responding ;-)

You don't necessarily have to take the derivative of bottom because it's a constant. Kind of like integral. Pull the constant out.

For instance, can you find the derivative of

$$\frac{\sin({\frac{x}{2} + \frac{1}{4}})}{\sin({1/4})}$$

It is the exact same idea, except y is your constant here.