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Problem Let y = ( x 2 + 1 ) sin x x 3 cos x y=(x^{2}+1)^{\sin x}\,x^{3\cos x} y = ( x 2 + 1 ) s i n x x 3 c o s x for x > 0 x>0 x > 0 . Which of the following equals 1 y d y d x \dfrac{1}{y}\dfrac{dy}{dx} y 1 d x d y ?
a) cos x ln ( x 2 + 1 ) + sin x x 2 + 1 − 3 sin x ln x + 3 cos x x \cos x\,\ln(x^{2}+1)+\dfrac{\sin x}{x^{2}+1}-3\sin x\,\ln x+\dfrac{3\cos x}{x} cos x ln ( x 2 + 1 ) + x 2 + 1 sin x − 3 sin x ln x + x 3 cos x
b) cos x ln ( x 2 + 1 ) + 2 x sin x x 2 + 1 − 3 sin x ln x + 3 cos x x \cos x\,\ln(x^{2}+1)+\dfrac{2x\sin x}{x^{2}+1}-3\sin x\,\ln x+\dfrac{3\cos x}{x} cos x ln ( x 2 + 1 ) + x 2 + 1 2 x sin x − 3 sin x ln x + x 3 cos x
c) cos x ln ( x 2 + 1 ) + 2 x sin x x 2 + 1 + 3 sin x ln x + 3 cos x x \cos x\,\ln(x^{2}+1)+\dfrac{2x\sin x}{x^{2}+1}+3\sin x\,\ln x+\dfrac{3\cos x}{x} cos x ln ( x 2 + 1 ) + x 2 + 1 2 x sin x + 3 sin x ln x + x 3 cos x
d) cos x ln ( x 2 + 1 ) + 2 x sin x x 2 + 1 − 3 sin x ln x − 3 cos x x \cos x\,\ln(x^{2}+1)+\dfrac{2x\sin x}{x^{2}+1}-3\sin x\,\ln x-\dfrac{3\cos x}{x} cos x ln ( x 2 + 1 ) + x 2 + 1 2 x sin x − 3 sin x ln x − x 3 cos x
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