39g'(x) = 1/f'( g(x) )
= 1/(2 + sin(g(x)) )
fog(x) = x
=> (2Ï€ - 3)3 +2g(x) + cos(g(x)) = x
How would we open the cosine and sine functions? I don't think the fof-1(x) property will come in use here...I get stuck at similar types of questions.
f[x]= [2Ï€-3]3 +2x-cos x Find.. ddxf-1[x]=? at x=Ï€ Note-[x] is not any greatest integer func.,just my keyboard wasn't working.
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4 Answers
39
62y= [2Ï€-3]3 +2x-cos x
We need to find the inverse .. so the simplest way is to replace x and y (of course there are lots of trivializations that i have made!)
x= [2Ï€-3]3 +2y-cos y
Where y is the inverse.. (though implicitly)
Now differentiate to find dy/dx