Asked by ACDub
derivative of h(x)= 3(√x) - x(√x)
Answers
Answered by
Reiny
I would rewrite as
h(x) = 3x^(1/2) - x^(3/2)
h ' (x) = (3/2) x^(-1/2) - (2/3) x^(1/2)
= 3/(2√x) - (2/3) √x
h(x) = 3x^(1/2) - x^(3/2)
h ' (x) = (3/2) x^(-1/2) - (2/3) x^(1/2)
= 3/(2√x) - (2/3) √x
Answered by
ACDub
Do you mean :
3/(2√x) - (3/2) √x. ?
3/(2√x) - (3/2) √x. ?
Answered by
Reiny
yes, of course, silly me, the exponent on the last one was 3/2
Answered by
ACDub
Is it required for me to use the chain rule on x(√x)?
Answered by
Reiny
it would be a combination of the chain rule and the product rule, doing it as a single power the way I did it was easier
as a product rule - chain rule for that term .....
x√x
= x x^(1/2)
derivative:
x (1/2) x^(-1/2) + (1)x^(1/2)
= x/(2√x) + √x
= x/(2√x) + 2x/√(2√x)
= 3x/(2√x)
= 3x/(2√x) * √x/√x , rationalizing the denominator
= 3x√x/(2x)
= 3√x/2 or (3/2)√x like I had before, but much much messier.
as a product rule - chain rule for that term .....
x√x
= x x^(1/2)
derivative:
x (1/2) x^(-1/2) + (1)x^(1/2)
= x/(2√x) + √x
= x/(2√x) + 2x/√(2√x)
= 3x/(2√x)
= 3x/(2√x) * √x/√x , rationalizing the denominator
= 3x√x/(2x)
= 3√x/2 or (3/2)√x like I had before, but much much messier.
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