Question

Find the derivatives

Answer 1

Challenge accepted!

Answer 2

1/sinx

Answer 3

x^2+x/sinx

Answer 4

-cosx/sin^2x; 2x+(sinx-xcosx)/(sin^2x)

Answer 5

(d/dx)f(g(x))=g'(x)f'(g(x)); thus, (d/dx)(1/sin(x))=-cos(x)/sin(x)^2
(d/dx)f(x)g(x)=f'(x)g(x)+g'(x)f(x); thus, (d/dx)(x^2+x/sin(x))=2x+(1/sin(x))-xcos(x)/sin(x)^2)

Answer 6

first one is cosecant and the derivative of cosecant is - cosecant cotangent

Answer 7

\[\sqrt{1-\sin^2x}\]

Answer 8

This is an application of the chain rule, Marashtii. I just answered a similar question earlier. Why don't you try doing it yourself, and telling me where you get confused?

Answer 9

\[\frac{1}{2\sqrt{1-\sin^2(x)}}\times -2\sin(x)\cos(x)\]

\[-\frac{\sin(x)\cos(x)}{|\cos(x)|}\]
so this is actually bit complicated. it will be sine if cosine is negative and minus sine if cosine is positive

Answer 10

we could have started with
\[\sqrt{1-\sin^2(x)}=|\cos(x)|\] and worked from there, again using the change rule and the fact that the derivative of
\[|x|\] is 1 if x is positive and -1 if x is negative

Answer 11

*chain rule

Answer 12

Actually, I did forget about the absolute value part.

/embarassed

Answer 13

Thanks, now it's clear :)