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f(x) = [5 \sqrt(x)] x [(x^3- (2 \sqrt(x) +2 ), find f'( x )?
derivatives. I don't even know which rule to use. I used the product rule and got 15x-6x^(-1/4)+5x^(-1/2)+2.5^(-3/2) but I'm wrong... anyone know it... after the formula I need to find f ' (x)=3 but the equation for the tangent is what I need....
I ma so confused..at this point I can't even remember the algebra that goes into it... I can't even remember if I can take 5 sqrt2 and muttiply it times 3x^2 and get 15x... is that even right? I had a 5 year gap between algebra and calc. I'm lost
2 Answers
- Anonymous10 years agoFavorite Answer
f(x) = 5√x * (x³ - 2√x - 2)
Use the product rule...
d/dx [fg] = f'g + fg'
We let...
f = 5√x = 5x^(½)
f' = 5/(2x^(½))
g = x³ - 2x^(½) - 2
g' = 3x² - 1/√(x)
So we have..
f'(x) = (5/(2√x))(x³ - 2x^(½) - 2) + 5√x * (3x² - 1/√x)
I hope this helps!
Source(s): Knowledge - Anonymous10 years ago
for this one you need to use the product rule and then the quotient rule.