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posted by Tony on Saturday, April 20, 2013 at 2:42pm.
If x=t^3 and y=3t^5, where t is any real number, what is the derivative of y with respect to x at x=125 ?
t=∛x =x^(1/3) => y=3x^5 = 3x^(5/3). dy/dx =3•(5/3)•x^(2/3) = =5 •125^(2/3)=525=125
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