CalculusHard ShareCalculus · Hard problemGiven siny=x\sin y=xsiny=x, find d2ydx2\dfrac{d^2y}{dx^2}dx2d2y in terms of xxx.A(1−x2)3/2x\dfrac{(1-x^2)^{3/2}}{x}x(1−x2)3/2Bx(1−x3)3/2\dfrac{x}{(1-x^{3})^{3/2}}(1−x3)3/2xCx(1−x2)3/3\dfrac{x}{(1-x^2)^{3/3}}(1−x2)3/3xDx(1−x2)3/2\dfrac{x}{(1-x^2)^{3/2}}(1−x2)3/2xCheck Show answer Walk through it step by step Practice more like this