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Problem Given h ( x ) = ( x 2 + 1 ) 3 cos ( x ) h(x)=(x^{2}+1)^{3}\cos(\sqrt{x}) h ( x ) = ( x 2 + 1 ) 3 cos ( x ) , which of the following is h ′ ( x ) h'(x) h ′ ( x ) ?
a) 6 x ( x 2 + 1 ) 2 cos ( x ) − ( x 2 + 1 ) 3 sin ( x ) 6x(x^{2}+1)^{2}\cos(\sqrt{x})-(x^{2}+1)^{3}\sin(\sqrt{x}) 6 x ( x 2 + 1 ) 2 cos ( x ) − ( x 2 + 1 ) 3 sin ( x )
b) 6 x ( x 2 + 1 ) 2 cos ( x ) − ( x 2 + 1 ) 3 sin ( x ) 2 x 6x(x^{2}+1)^{2}\cos(\sqrt{x})-\dfrac{(x^{2}+1)^{3}\sin(\sqrt{x})}{2\sqrt{x}} 6 x ( x 2 + 1 ) 2 cos ( x ) − 2 x ( x 2 + 1 ) 3 sin ( x )
c) 6 x ( x 2 + 1 ) 2 cos ( x ) − ( x 2 + 1 ) 3 cos ( x ) 2 x 6x(x^{2}+1)^{2}\cos(\sqrt{x})-\dfrac{(x^{2}+1)^{3}\cos(\sqrt{x})}{2\sqrt{x}} 6 x ( x 2 + 1 ) 2 cos ( x ) − 2 x ( x 2 + 1 ) 3 cos ( x )
d) 3 ( x 2 + 1 ) 2 cos ( x ) − ( x 2 + 1 ) 3 sin ( x ) 2 x 3(x^{2}+1)^{2}\cos(\sqrt{x})-\dfrac{(x^{2}+1)^{3}\sin(\sqrt{x})}{2\sqrt{x}} 3 ( x 2 + 1 ) 2 cos ( x ) − 2 x ( x 2 + 1 ) 3 sin ( x )
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