Differentiate d/dx(x3.sin4x)
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Payal Singh 7 years ago
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}
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