find the derivatives of the following
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(x + 1/x)^3 = x^3 + 1/x^3 + 3x + 3/x
=d/dx (x^3 + 1/x^3 + 3x + 3/x)
=d/dx(x^3 + x^-3 + 3x + 3x^-1)
= 3x^2 - 3x^-4 + 3 - 3x^-2
Thanks
=d/dx (x^3 + 1/x^3 + 3x + 3/x)
=d/dx(x^3 + x^-3 + 3x + 3x^-1)
= 3x^2 - 3x^-4 + 3 - 3x^-2
Thanks
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