B bluboy22 New Member Joined Aug 16, 2009 Messages 8 Gender Undisclosed HSC N/A Aug 16, 2009 #1 Find dy/dx using the product and chain rule y= x^2(x-5)^1/4 TIA
lychnobity Active Member Joined Mar 9, 2008 Messages 1,292 Gender Undisclosed HSC 2009 Aug 16, 2009 #2 bluboy22 said: Find dy/dx using the product and chain rule y= x^2(x-5)^1/4 TIA Click to expand... y' = 2x(x-5)1/4 + (x2)0.25(x-5)-3/4
bluboy22 said: Find dy/dx using the product and chain rule y= x^2(x-5)^1/4 TIA Click to expand... y' = 2x(x-5)1/4 + (x2)0.25(x-5)-3/4
B bluboy22 New Member Joined Aug 16, 2009 Messages 8 Gender Undisclosed HSC N/A Aug 16, 2009 #3 lychnobity said: y' = 2x(x-5)1/4 + (x2)0.25(x-5)-3/4 Click to expand... Thanks, but was the chain rule used as well to get this answer?
lychnobity said: y' = 2x(x-5)1/4 + (x2)0.25(x-5)-3/4 Click to expand... Thanks, but was the chain rule used as well to get this answer?
lychnobity Active Member Joined Mar 9, 2008 Messages 1,292 Gender Undisclosed HSC 2009 Aug 16, 2009 #4 bluboy22 said: Thanks, but was the chain rule used as well to get this answer? Click to expand... Yeah, sorry for the rushed answer. In particular, when differentiating (x-5)<sup>1/4</sup> the chain rule has to be used. ie the 2nd half of the answer (x<sup>2</sup>)0.25(x-5)<sup>-3/4</sup>
bluboy22 said: Thanks, but was the chain rule used as well to get this answer? Click to expand... Yeah, sorry for the rushed answer. In particular, when differentiating (x-5)<sup>1/4</sup> the chain rule has to be used. ie the 2nd half of the answer (x<sup>2</sup>)0.25(x-5)<sup>-3/4</sup>