Question

can anyone pls answer this?
solve and simplify the derivative of
cot^2(y+2x)+ln[csc(y+2x)] = cos pi

Answer 1

@lgbasallote i think your smart enough to solve this can you solve this for me?

Answer 2

so you're supposed to take the derivative of that?

Answer 3

yes @lgbasallote

Answer 4

\[\frac{\cos^2(y+2x)}{\sin^2(y+2x)}+\ln(\frac{1}{\sin(y+2x)})=\cos(\pi)=-1\]

Answer 5

First- use chain rule and differentiate wrt (y+2x).

-For the cotangent use the product rule for the denominator and numerator
-For the natural log use the chain rule a couple of times.

Answer 6

Do you have any idea how to start on this?

Answer 7

@happy87 ?

Answer 8

yes @henpen i'm sorry for the late reply

Answer 9

can you show me the whole solution @henpen ? thanks that would be very helpful

Answer 10

Differenitate both sides wrt t: -1 turns to 0.

Answer 11

Addition rule or whatever it's called- the two terms on each side of the + (or -) differentiate independently.

Answer 12

Let's work on the right one first.

Answer 13

\[\frac{dln\frac{1}{\sin(y+2x)}}{dx}= \frac{dln\frac{1}{\sin(y+2x)}}{d\frac{1}{\sin(y+2x)}} \frac{ d\frac{1}{\sin(y+2x)} }{dsin(y+2x)}\frac{dsin(y+2x)}{d(y+2x)}\frac{d(y+2x)}{dx}\]

Answer 14

So, for example, treat the first term in the multiplication as