anonymous
  • anonymous
Derived \[\sin (\frac{ x+1 }{ 2x-3 } )\]
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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UnkleRhaukus
  • UnkleRhaukus
do you remember the chain rule ?
anonymous
  • anonymous
no
UnkleRhaukus
  • UnkleRhaukus
\[(f\circ g)'(t)=f'(g(t))\cdot g'(t)\]

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UnkleRhaukus
  • UnkleRhaukus
where \(f\circ g=f(g(t))\)
UnkleRhaukus
  • UnkleRhaukus
in your question f =sin
UnkleRhaukus
  • UnkleRhaukus
can you tell me what g(x) is ?
anonymous
  • anonymous
sin
UnkleRhaukus
  • UnkleRhaukus
nope,
anonymous
  • anonymous
then .. can u solve this derived
UnkleRhaukus
  • UnkleRhaukus
To take the derivative of the function-of-a-function, you have to be able to identify the functions
UnkleRhaukus
  • UnkleRhaukus
your function is \[y=\sin\left(\frac{x+1}{2x−3}\right)\] you need to decompose \(y=f(g(x))\) \[f=\sin \\\, \\\,\\ g(x)=?\]
anonymous
  • anonymous
x+1/2x-3
UnkleRhaukus
  • UnkleRhaukus
ok, so now look at the rule for takig the derivative of a function-of-a-function \[(f(g(t)))'=(f∘g)'(t)=f'(g(t))⋅g'(t)\]
UnkleRhaukus
  • UnkleRhaukus
you need to take the derivative of f, \[f=\sin\\f'= \]
anonymous
  • anonymous
cos
UnkleRhaukus
  • UnkleRhaukus
correct ,
UnkleRhaukus
  • UnkleRhaukus
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 \[y=\frac{u}{v}\\y'=\frac{vu'-uv'}{v^2}\]
UnkleRhaukus
  • UnkleRhaukus
so you will have to decompose \[\frac{x+1}{2x−3}\] onto \(\dfrac uv\) form and then find \(u,u',v,v',v^2\)
anonymous
  • anonymous
nd cos ..?
UnkleRhaukus
  • UnkleRhaukus
pardon
anonymous
  • anonymous
no pasa nada
anonymous
  • anonymous
np prob
UnkleRhaukus
  • UnkleRhaukus
do you still need any help on this question?
anonymous
  • anonymous
yes

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