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Please help.Find dy/dx when:?
f(x)=sin(logx) and y=f[(2x+3)/(3-2x)]
2 Answers
- SidLv 61 decade agoFavorite Answer
f(x) = sin(logx) and y = f[(2x + 3)/(3 - 2x)]
Since f(x) = sin(logx), so,
y = f[(2x + 3)/(3 - 2x)] = sin[ log[(2x + 3)/(3 - 2x)] ]
dy/dx = cos[ log[(2x + 3)/(3 - 2x)] ] * d/dx [log[(2x + 3)/(3 - 2x)]]
dy/dx = cos[ log[(2x + 3)/(3 - 2x)] ] * [(3 - 2x)/ (2x + 3)] * d/dx [(2x + 3)/(3 - 2x)]
dy/dx = cos[ log[(2x + 3)/(3 - 2x)] ] * [(3 - 2x)/ (2x + 3)] * [12/(3 - 2x)^2]
- 1 decade ago
Hi,
We are given the following information :
f(x)= sin ( log x)
y=f[(2x+3)/(3-2x)]
Therefore y= sin ( log[(2x+3)/(3-2x)] )
Differentiating w.r.t x, we obtain the following :
dy/dx= {cos ( log[(2x+3)/(3-2x)] ) }*{(3-2x)/(2x+3)}*{[ 2(3-2x) + 2(2x+3) ]/(3-2x)^2 }
= [12/(3-2x)(2x+3)] * cos( log[(2x+3)/(3-2x)] )
I hope this helps
Have fun!
Regards
