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Find dy/dx:Y = \(\cfrac{x^2\,sin\,x}{(1+x^2)^2\sqrt{1+x}}\) |
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Answer» Y = \(\cfrac{x^2\,sin\,x}{(1+x^2)^2\sqrt{1+x}}\) \(\therefore\) \(\cfrac{dy}{dx}\) = \(\frac{(1+x^2)^2\sqrt{1+x}\frac{d}{dx}(x^2sin\,x)-x^2\,sin\,x\frac{d}{dx}[(1+x^2)^2\sqrt{1+x}]}{(1+x^2)^4(1+x)}\) = (1 + x2)2 \(\sqrt{1+x}\) (x2 cos x + 2 x sin x) - \(\frac{x^2\,sin\,x((\frac{(1+x^2)^2}{2\sqrt{1+x}}+4x(1+x^2)\sqrt{1+x})}{(1+x^2)^4(1+x)}\) = \(\frac{2(1+x^2)^2({1+x)}(x^2cos\,x+2x\,sin\,x)]-x^2sin\,x((1+x^2)^2+8x(1+x)(1+x^2))}{(1+x^2)^4(1+x)}\) = \(\frac{x(1+x^2)2(1+x^2)(1+x)(x\,cos\,x+2sinx)-xsinx(1+x^2+8x(1+x)}{(1+x^2)^4)(1+x)}\) = \(\frac{x(2(1+x^2)(1+x)(x\,cosx+2sinx)-xsinx(9x^2+8x+1))}{(1+x^2)^3(1+x)}\) |
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