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1. (3x^2 +2x-1) ^1/2 (square root, but i can't express it)
_________________
2^x +x

....i go to use the quotient rule but i'm not sure if you have to factor the top at some point or what happens with the variable in 2^x

2. (ax+b) (x^3 + 1/2x^2 +2logx) where a and b are constants.
....no idea on this one.

3. 1/ 3(2x-1)^2 +2^5x
....what happends with 2^5x, do you just find the derivative of the first part thenadd its derivative on to that?


Thank you, I appreciate this more than you'll ever know!!!

2007-11-04 05:30:52 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

When applying rules for derivatives, you have to be very methodical:

#1. (f(x)/g(x))' = (f'(x)g(x)-f(x)g'(x))/(g(x)^2)

In this case, f(x) = (3x^2 +2x-1) ^0.5 and g(x) = (3x^2 +2x-1) ^1/2

Hint: to compute f'(x), you need to use the rules:
(f(g(x)))' = f'(g(x))*g'(x) and (f(x) + g(x))' = f'(x) + g'(x)

f'(x) = 0.5*((3x^2 +2x-1)^-0.5)*(6x +2)

Hint: to compute g(x), you need to use the rule:
(f(x) + g(x))' = f'(x) + g'(x)

g'(x) = (2^x)*ln(2) + 1

Now, you have f(x), g(x), f'(x) and g'(x), plug them in the formula I put on top of #1 to get your result.

#2. (f(x)*g(x))' = f'(x)*g(x) + f(x)*g'(x)

In this case, f(x) = (ax+b) and g(x) = (x^3 + 1/2x^2 +2logx)

f'(x) = a
g'(x) = 3*(x^2) + x + 2/(x*ln10)

Now, you have f(x), g(x), f'(x) and g'(x), plug them in the formula I put on top of #2 to get your result.

#3. (f(x) + g(x))' = f'(x) + g'(x)

Here, f(x) = 1/ (3(2x-1)^2) and g(x) = 2^(5x)

Hint: to compute f'(x), you need to use the rule: (f(x)/g(x))' = (f'(x)*g(x)-g'(x)*f(x))/(g(x)^2)

f'(x) = (-12*(2x-1))/(9*((2x-1)^2))
g'(x) = 5*(2^(5x))*ln2

Now, you have f(x), g(x), f'(x) and g'(x), plug them in the formula I put on top of #3 to get your result.

If you want all the formulas for the derivatives, refer to:
http://en.wikipedia.org/wiki/Table_of_derivatives

Just remember that you might have to use several of the formulas just to solve one derivative.

2007-11-04 06:13:07 · answer #1 · answered by Ricky V 1 · 0 0

1. derivative of numerator = (6x+2)/(3x^2+2x-1)^.5
derivative of denominator is 2^xln(2) +1
Now you can use quotient rule.

2. Use product rule
The derivative of (ax+b) = a
Derivative of (x^3+1/2 x^2 +2logx) is 3x^2 +x +2/x

3. derivative of 2^5x = (2^5xln(2))(5)

2007-11-04 05:44:56 · answer #2 · answered by ironduke8159 7 · 0 0

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2016-11-10 06:17:45 · answer #3 · answered by larrinaga 4 · 0 0

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