derivitive + power?

Question

finding the derivitive of 2s^(3/2)

so far i've done (3/2) - 1 = 1/2

my answer is s^(1/2) but the answer to the problem is 3s^(1/2)..i'm not sure where the 3 comes from

Answer 1

You just have to believe and memorize the rule correctly. However, if you only base on knowing the rule, then you can't check yourself if you had remembered correctly.

I always suggest knowing the fundamental, and the mathematical way. Many "mathematical" rules such as this derivative rule are not mathematical, they are only pattern re-arrangment, or to people who don't want to understand the mathematics behind them, they are just magic.

Derivative is related to gradient, or rate of change of the expression with respect to change in one of its variable.

Using a small change in the variable, delta_x, we may approximate the derivative by
(f(x+delta_x) - f(x)) / (x+delta_x - x)
= (f(x+delta_x) - f(x)) / delta_x

The smaller the delta_x, the more accurate the above approximate for gradient. In calculus, we consider delta_x just bigger than 0:
limit of (f(x+deltax) - f(x)) / deltax as delta_x --> 0

For this case where f(x) = 2x^(3/2),
f'(x)
= limit of (f(x+delta_x) - f(x)) / deltax as delta_x --> 0
The numerator of the main term: (f(x+delta_x) - f(x))
= 2(x+delta_x)^(3/2) - 2x^(3/2)
= 2(x^3/2 + ((3/2)/1!)x^(3/2-1)delta_x + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x^2 + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^n) - 2x^(3/2) + ...
= 2( (x^3/2 + ((3/2)/1!)x^(3/2-1)delta_x + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x^2 + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^n) + ... ) - x^(3/2) )
= 2( ((3/2)/1!)x^(3/2-1)delta_x + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x^2 + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^n) + ...)
Now, divide the above by deltax:
= 2( ((3/2)/1!)x^(3/2-1) + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^(n-1)) + ... )

Thus, when delta_x is almost zero, or at zero,
limit of (f(x+delta_x) - f(x)) / deltax as delta_x --> 0
= limit 2( ((3/2)/1!)x^(3/2-1) + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^(n-1)) + ... ) as delta_x --> 0
= 2( ((3/2)/1!)x^(3/2-1) + ((3/2)(3/2-1)/2!)x^(3/2-2)delta_x + .. + ((3/2)(3/2-1)..(3/2-(n-1))/n!)(delta_x)^(n-1)) + ... ) when delta_x = 0
= 2( ((3/2)/1!)x^(3/2-1) + 0 + .. + 0 + ... )
= 2( ((3/2)/1!)x^(3/2-1) )

Thus, from here, we knows that the coefficient is not affected, the power happens to be the coefficient of the 2nd term in the binomial expansion, this 2nd term has the power of x smaller than the original power by 1, the first term was subtracted away, and all the other terms having delta_x in them become zero.

This works for any ax^n terms, where the above binomial expansion is valid. Thus, substituting 2 and 3/2 with a and n gives a( ((n)/1!)x^(n-1) ) = anx^(n-1)

I suggest to use the the above mathematical story to help remember the magical derivative rule.

Answer 2

2 * 3/2 *s^(3/2-1) = 3 s^(1/2)

Answer 3

"Multiply the variable by the power and subtract 1 from the power."
f(s) = 2s^(3/2)
f'(s) = (3/2)(2)s^(3/2-2/2)
f'(s) = 3s^1/2

Answer 4

The power rule says to take the exponent, multiply it by the coefficient to get the new coefficient, then subtract one from the exponent to get the new exponent. In other words, the step you missed is to multiply 2 by 3/2 to get 3.