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| 1. |
Differentiated/dx(x3.sin4x)\xa0 |
| Answer» Use chain rule for it,{tex}{d\\over dx}(x^3.sin4x){/tex}{tex}= x^3.{d\\over dx}(sin 4x) + sin4x. {d\\over dx} (x^3){/tex}{tex}= x^3.cos4x.{d\\over dx}(4x) + sin4x.3x^2{/tex}{tex}= 4x^3cos4x + 3x^2sin4x{/tex} | |