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find dy/dx if Y=u^1/3 and u=x^4-4x^3-4. I get 1/3(4x^7-28x^6+48x^5-16x^3+48x^2) This does not seem like it is correct to me.... What do you think?
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muraad come to my question
no...answer is \[\left(4x ^{3}-12x ^{2} \right)/\left( 3\left( x ^{4}-4x ^{3}-4 \right)^{1/2} \right)\]
I got to the point of 1/3(x^4-4x^3-4)^-2/3 (4x^3-12x^2) for your answer I understand the (4x^3-12x^2) but for the bottom what happened to the ^-2/3? in 1/3(x^4-4x^3-4)^ -2/3
u^(-2/3) means 1/[u^(2/3)]. x^-2 means 1/(x^2).
oh...sorry...I made a mistake...i thought the exponent was 1/2...ill recalculate...gimme a sec
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ok...the answer is the same...just the exponent in the denominator turns to 2/3..not 1/2
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