Question

Найдите производнуюю плиииз

Answer 1

$y'=(tg(ln(arcsin(2^{\sqrt{sinx}}))))'=\\=\frac{1}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))}*(ln(arcsin(2^{\sqrt{sinx}})))'=\\=\frac{1}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))}*\frac{1}{arcsin(2^{\sqrt{sinx}})}*(arcsin(2^{\sqrt{sinx}}))'=\\=\frac{1}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))arcsin(2^{\sqrt{sinx}})}*\frac{1}{\sqrt{1-(2^{\sqrt{sinx}})^2}}*(2^{\sqrt{sinx}})'=$
$=\frac{1}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))arcsin(2^{\sqrt{sinx}})\sqrt{1-(2^{\sqrt{sinx}})^2}}*ln2*2^{\sqrt{sinx}}*\\ *(\sqrt{sinx})'=\\=\frac{ln2*2^{\sqrt{sinx}}}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))arcsin(2^{\sqrt{sinx}})\sqrt{1-(2^{\sqrt{sinx}})^2}}*\frac{1}{2\sqrt{sinx}}*(sinx)'=\\=\frac{cosx*ln2*2^{\sqrt{sinx}}}{cos^2(ln(arcsin(2^{\sqrt{sinx}})))arcsin(2^{\sqrt{sinx}})\sqrt{1-(2^{\sqrt{sinx}})^2}*2\sqrt{sinx}}$