Question

derived...
f(x)=sin \left( x+1/2x-3 \right)

Answer 1

\[f(x)=\sin \left( x+1/2x-3 \right) \]

Answer 2

yess

Answer 3

You want to find its derivative ??

Answer 4

yess

Answer 5

See here:

\[\large \frac{d}{dx}(\sin(f(x)) = \cos(f(x)) \times \frac{d}{dx}(f(x))\]

Answer 6

So here:

What is the term in place of f(x) in sin(f(x)) ??

Answer 7

sin has its derivative as cos..

So we are left with the derivative of f(x)

That will be:

\[\frac{d}{dx}(\sin(\frac{x+1}{2x-3})) = \cos(\frac{x+1}{2x-3}) \times \frac{d}{dx}(\frac{x+1}{2x-3})\]

Answer 8

so by using division rule you can find the derivative of last term..

Answer 9

no,,

Answer 10

\[\large \frac{u}{v} = \frac{v \cdot u' - u \cdot v'}{v^2}\]

Answer 11

\[\frac{d}{dx}(\frac{x+1}{2x-3}) = \frac{(2x-3)(1) - (x+1)(2)}{(2x-3)^2} \implies \frac{-5}{(2x-3)^2}\]

Answer 12

Now combine them:

\[\frac{d}{dx}(\sin(\frac{x+1}{2x-3})) = \cos(\frac{x+1}{2x-3}) \times \frac{-5}{(2x-3)^2}\]

Answer 13

f(x)=sin(x+1/2x−3)

use chain rule

g(x) = sin(x)
g'(x) = cos(x)
f(x) = (x+1)/(2x-3)
f'(x) = (see below)

Deriving f'(x):
s(x) = x+1
s'(x) = 1

m(x) = 2x-3
m'(x) = 2

we use

\[\frac{s'(x)m(x) - s(x)m'(x)}{(m(x))^2}\]

so we have

f'(x) = \[\frac{(2x-3) - (2)(x+1)}{(2x-3)^2}\]

Now we just use chain rule

g(x) = sin(x)
g'(x) = cos(x)
f(x) = (x+1)/(2x-3)
f'(x) = \[\frac{(2x-3) - (2)(x+1)}{(2x-3)^2}\]


So knowing this we apply

g'(f(x)) * f'(x)

so

\[\cos(\frac{x+1}{2x-3})*(\frac{(2x-3) - (2)(x+1)}{(2x-3)^2})\]

Answer 14

the bottom equation is f'(x), also I made a mistake and used f(x) in the solution I should have used like l(x) or something for (x+1)/(2x-3)

so just pretend
f(x) is l(x)
so,
l(x) = (x+1)/(2x-3)

to avoid confusion

Answer 15

the best way to deal with derivatives when you first start out I find is to just split them up into separate functions, take the derivative of those smaller functions and then stick them back into the applicable formula to get the final answer

Answer 16

also note i split l(x) into two functions so

l(x) = s(x)/m(x)

Answer 17

if you have any questions feel free to ask, really once you do it this way derivatives are super easy eventually you won't even need to split them up, you can just do them in your head super quick

Answer 18

can u give me all formulas the derived

Answer 19

second page of this document

Answer 20

ok thanks

Answer 21

to be honest you don't have to memorize quotient rule, you can just use product rule with chain rule

\[\frac{(x+1)}{(2x-1)} = (x+1)(2x-1)^{-1}\]

Answer 22

but that is just a note for the future :)

Answer 23

uv=v⋅u′−u⋅v′v2

u means??

Answer 24

you would just have

q(x) = x+1
q'(x) = 1

d(x) = (2x - 1)^(-1) use chain rule here
d'(x) = -1(2x-1)^(-2) * 2

then just plug into product rule

q'(x)d(x) + d'(x)q(x)

Answer 25

but if you are just starting with derivatives its probably best you learn quotient

Answer 26

hey!
put your equation in 'WOLFramalpha'
and u can get the result
its an outstanding computational website
just put ur equation in its search engine and it will generate output of ur equation!!!!@Muskan

Answer 27

http://www.wolframalpha.com/

Answer 28

yeah wolfram alpha can be good to check your work but sometimes your answers can end up looking different even though they are correct and the same so just be aware

Answer 29

@Australopithecus i put above equation in it and it give a correct answer with graph
i love it(wolfram alpha)

Answer 30

you should advertise that site for a living

Answer 31

@Australopithecus it makes work load easy!

Answer 32

it is a nys site
thanks @ali110

Answer 33

fundamental knowledge is the key to solving "difficult" questions