can some one help please?

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mathhelp1a

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Oct 4, 2009
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40
differentiate and simplify as indicated

y= 3/ sqrt x^2

is it -3x / x^2

is that right
 
y = f(x) = 3(x2) = 3∣x∣\displaystyle y \ = \ f(x) \ = \ \frac{3}{\sqrt (x^{2})} \ = \ \frac{3}{|x|}

Hence, f′(x) = −3x∣x∣\displaystyle Hence, \ f'(x) \ = \ \frac{-3}{x|x|}
 
so the answer is -3 / absoulute x^2
how did sqrt x get changed to abosoulte x ???
 
Hey, y=3/sqrt x2 is y = 3(x2) is 3∣x∣.\displaystyle Hey, \ y= 3/ sqrt \ x^2 \ is \ y \ = \ \frac{3}{\sqrt(x^{2})} \ is \ \frac{3}{|x|}.
 
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