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Can anyone help me differentiate this: 1/3 x^3 +3/2x^2-7
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Hint: \[\frac{d}{dx} x^n=nx^{n-1}\]\[\frac{d}{dx} C\cdot f(x) = C\cdot \frac{d}{dx}f(x)\]\[\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)\]
is that ( 1/(3x^(3)) ) + ( 3/(2x^(2)-7) )
well they are added s just differentiate ( 1/(3x^(3)) ) and ( 3/(2x^(2)-7) ) then add the two derivatives 1/(3x^(3)) = 3x^(-3) and ( 3/(2x^(2)-7) )= 3(2x^(2)-7)^(-1) thus we have 3x^(-3) + 3(2x^(2)-7)^(-1) can you solve this?
for the first one use the rule yakeyglee posted for the second one use chain rule
Note I simply used this rule to put it in a form that is easier to deal with |dw:1339004017955:dw|
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ok I get it
thanks everyone
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