Question

Derived
\[\sin (\frac{ x+1 }{ 2x-3 } )\]

Answer 1

do you remember the chain rule ?

Answer 2

no

Answer 3

\[(f\circ g)'(t)=f'(g(t))\cdot g'(t)\]

Answer 4

where \(f\circ g=f(g(t))\)

Answer 5

in your question

f =sin

Answer 6

can you tell me what g(x) is ?

Answer 7

sin

Answer 8

nope,

Answer 9

then ..
can u solve this derived

Answer 10

To take the derivative of the function-of-a-function, you have to be able to identify the functions

Answer 11

your function is

\[y=\sin\left(\frac{x+1}{2x−3}\right)\]

you need to decompose \(y=f(g(x))\)

\[f=\sin \\\,
\\\,\\
g(x)=?\]

Answer 12

x+1/2x-3

Answer 13

ok, so now look at the rule for takig the derivative of a function-of-a-function

\[(f(g(t)))'=(f∘g)'(t)=f'(g(t))⋅g'(t)\]

Answer 14

you need to take the derivative of f,

\[f=\sin\\f'= \]

Answer 15

cos

Answer 16

correct ,

Answer 17

now all you need before you can subsitut into that formula is to find

\[g'(t)\]


this is a bit tricker , and you might want to use the quotient rule

\[y=\frac{u}{v}\\y'=\frac{vu'-uv'}{v^2}\]

Answer 18

so you will have to decompose

\[\frac{x+1}{2x−3}\]

onto \(\dfrac uv\) form

and then find \(u,u',v,v',v^2\)

Answer 19

nd cos ..?

Answer 20

pardon

Answer 21

no pasa nada

Answer 22

np prob

Answer 23

do you still need any help on this question?

Answer 24

yes