find the first derivative y=(csc x + cot)(csc x - cot)
Do you know the product rule?
yeah
Do you know the derivatives of those trig functions? :)
no
The derivative this time is 0
y = (csc x + cot x)(csc x - cot x) = (csc x)^2 - (cotx)^2 = 1 + (cotx)^2 - (cotx)^2 =1 And the derivative of 1 is 0
Maybe he'd like to know how to do it instead of just being given the answer?
how you get the power to the 2 for csc x and cot x?
By simplifying the expression, multiplying what's inside the parenthesis
so i have to use foil?
Yep, use the foil method
ok. i dont understand the csc and cot like what do they equal to?
What do you mean by that? Do you mean what they equal after you multiply them?
like sin x= cos x and cos x = -sinx or tan x= sin x/cos x
csc x = 1/sin x and tan = 1/tan x
No no no! The derivative of csc x is -cosxcotx and the derivative of tan x is sec^2 x. While yes, you can foil this before you find the derivative, you don't have to.
Oh, I got confused for a second there...oops ^^*
Becki is right, but using foil just simplifies the expression to cut back on confusion
so csc x = -cosx cot x and tan x = sec^2? What about sin and cos?
sin x and cos x you had correct. the deriviatives of each of those is cos x and -sin x respectively.
so they equal to each other?
No. When you differentiate sin x, you get cos x. When you differentiate cos x, you get -sin x.
oooh ok but what about dividing cos x and sin x?
Dividing them by what?
cos x / sin x
That's the definition of cot x.
i though it equal to tan x
Here's what i got for your original question. I went ahead and foiled it since you probably already did that. Then I used the chain rule to differentiate.
how you get (2cot x)? i though the 2 supposed to be on the outside of the parentheses
never mind
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