Question

Differentiate the identity sin 2x = 2sin x cos x to develop the identity for cos 2x , in terms of
sin x and cos x.

Answer 1

like what are they trying to ask? i dont get it

Answer 2

take the derivatives of sin(2x) and 2sin x cos x, then set the two equal to each other and solve for cos(2x)

Answer 3

in terms of sin x and cos x

Answer 4

ok so when i take the derv of sin2x i get 2sinxcosx and the right side is the same ? doesnt make seice

Answer 5

take the derivative of sin2x using the chain rule

Answer 6

sin2x = 2sinxcosx; the derivative of sin2x is not equal to 2sinxcosx

Answer 7

oohh damm yeea its 2cos2x am i right?

Answer 8

yup :) now take the derivative of 2sinxcosx using the product rule

Answer 9

would it be -cosxsinx??

Answer 10

let f(x)=2sin(x) and g(x)=cos(x)
the product rule states that the derivative of f(x)g(x) is equal to f(x)g'(x)+f'(x)g(x)

Answer 11

ok so i get -2sin^(2)x + cos^(2)x

Answer 12

close. for the second term, remember to carry over the 2 when taking the derivative of 2sin(x)

Answer 13

like?

Answer 14

f'(x) * g(x) = [2cos(x)] * [cos(x)] = 2 cos^(2)x

Answer 15

ohh yeaa

Answer 16

now what do i do ?

Answer 17

set the two derivatives equal to each since the functions that we differentiated (sin(x) and 2sin(x)cos(x)) are equal to each other

Answer 18

and then solve for cos(2x)

Answer 19

2cos2x = -2sin^(2)x + 2cos^(2)x
thus cos(2x) = cos^(2)x-sin^(2)x

Answer 20

hey here is my new queston. differentitiate the function y=1-cosx/sinx and y=cscx- cotx , and show that the derivaties are equal

Answer 21

so i toke the derv of the 2ns eqaution and i get the first equation but they say the derv habve to be the same so like what ?

Answer 22

do i take the derv of them both and they have to end up the same ? ?

Answer 23

yup, the derivatives should end up the same. also the first function is (1-cosx)/sinx

Answer 24

can u help me take the derv of the first function , yea ur right

Answer 25

to take the derivative of (1-cosx)/sinx, use the quotient rule. f(x)=(1-cosx), g(x)=sinx, and the derivative of f/g = (f'(x)g(x)-f(x)g'(x)) / (g(x)^2)

Answer 26

http://www.1728.org/chainrul.htm

Answer 27

Ok so this is what i get

Answer 28

-sin^(2)x-cosx+cox^(2)x/(sinx)^2 right?

Answer 29

looks good except for the first term. be careful when taking the derivative of 1-cos(x)
f'(x) = d/dx (1) + d/dx (-cos(x)) = - d/dx (cos(x)) = sin(x)
f'(x)g(x) = sin^(2)x

Answer 30

ok so then top i get 1-cosx looks good but for my botton its sin^(2)x it should be just sinx thoug

Answer 31

sin^(2)x is good for the denominator

Answer 32

but if u take the derv of the 2nd funstion its just sinx for the deno

Answer 33

if you're allowed to use a trig derivative table
d/dx (cscx - cotx) = [-csc(x)cot(x)] - [-csc(x)^2]
otherwise, you can use the quotient rule to find the derivatives of csc(x) and cot(x)

Answer 34

the derv for cscx is 1/sinx and for cotx its cosx/sinx

Answer 35

cscx = 1/sin(x) and cot(x)=cos(x)/sin(x)

Answer 36

so it would be 1/sinx-cox/sinx right

Answer 37

so becasue the deno is the same it would be 1-cosx/sinx right ?

Answer 38

you need to find the derivatives though. either use a table, or use the quotient rule to find the derivative of 1/sin(x) and the derivative of cos(x)/sin(x)

Answer 39

dammnn yea

Answer 40

csc(x)-cot(x)=1/sinx-cox/sinx, since by definition, csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x)
here is a trig table: http://sub.allaboutcircuits.com/images/11045.png

Answer 41

to manually find the derivative of csc(x), which equals 1/sin(x), let f(x)=1 and g(x)=sin(x), then use the quotient rule

Answer 42

ok man thanks so much

Answer 43

glad to have helped. let me know if you have any other questions :)

Answer 44

oo yea i have one more

Answer 45

here is another one , if f(x)=sinx,evaluate lim h-->0 f(2+h)-f(2)/h , to two decimel places

Answer 46

^ the question