A) \[\frac{xyz}{t}\]
B) \[2\frac{xyz}{t}\cos ({{x}^{2}})\]
C) \[\frac{-xyz\,cos({{x}^{2}})}{t}\]
D) \[\frac{xyz\,t}{\cos ({{x}^{2}})}\]
Correct Answer: B
Solution :
Given, \[y=\sin {{x}^{2}},z={{e}^{{{y}^{2}}}},t=\sqrt{z}\] Now, \[\frac{dy}{dx}=\frac{d\sin {{x}^{2}}}{dx}=\cos ({{x}^{2}}).2x\] \[\therefore \] \[\frac{dz}{dy}=\frac{d{{e}^{{{y}^{2}}}}}{dy}={{e}^{{{y}^{2}}}}.2y\] \[\frac{dt}{dz}=\frac{d\sqrt{z}}{dz}=\frac{1}{2\sqrt{z}}\] \[\therefore \] \[\frac{dt}{dx}=\frac{dt}{dz}\times \frac{dz}{dy}\times \frac{dy}{dx}\] \[=\frac{1}{2\sqrt{z}}\times {{e}^{{{y}^{2}}}}\times 2y\times 2x\times \cos ({{x}^{2}})\] \[=\frac{2xyz}{t}.\cos ({{x}^{2}})\]
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