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Question Number 166585 by mkam last updated on 22/Feb/22
if y = (sinx)^e^x   find ((d ( e^(sin^(−1) y) ))/(d (lnx)))
$${if}\:{y}\:=\:\left({sinx}\right)^{{e}^{{x}} } \:{find}\:\frac{{d}\:\left(\:{e}^{{sin}^{−\mathrm{1}} {y}} \right)}{{d}\:\left({lnx}\right)} \\ $$
Answered by Eric002 last updated on 22/Feb/22
ln(y)=e^x  ln(sin(x))  (dy/dx)=e^x (sin(x))^e^x  )(cot(x)+ln(sin(x))  d(e^(sin^(−1) (y)) )/d(ln(x))=((x e^(sin^(−1) (y)) /(√(1−y^2 )))/dx)dy   =e^x (sin(x))^e^x  (cot(x)+ln(sin(x)))(((xe^(sin^1 (y)) )/( (√(1−y^2 )))))
$${ln}\left({y}\right)={e}^{{x}} \:{ln}\left({sin}\left({x}\right)\right) \\ $$$$\left.\frac{{dy}}{{dx}}={e}^{{x}} \left({sin}\left({x}\right)\right)^{{e}^{{x}} } \right)\left({cot}\left({x}\right)+{ln}\left({sin}\left({x}\right)\right)\right. \\ $$$${d}\left({e}^{{sin}^{−\mathrm{1}} \left({y}\right)} \right)/{d}\left({ln}\left({x}\right)\right)=\frac{{x}\:{e}^{{sin}^{−\mathrm{1}} \left({y}\right)} /\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{{dx}}{dy}\: \\ $$$$={e}^{{x}} \left({sin}\left({x}\right)\right)^{{e}^{{x}} } \left({cot}\left({x}\right)+{ln}\left({sin}\left({x}\right)\right)\right)\left(\frac{{xe}^{{sin}^{\mathrm{1}} \left({y}\right)} }{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\right) \\ $$

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