derivative of h(x)= 3(√x) - x(√x)
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
Do you mean :
3/(2√x) - (3/2) √x. ?
yes, of course, silly me, the exponent on the last one was 3/2
Is it required for me to use the chain rule on x(√x)?
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.
To find the derivative of the function h(x) = 3√x - x√x, we need to use the rules of differentiation. The function h(x) can be expressed as a sum or difference of terms, where each term is a product or a composition of functions.
Let's break down the function h(x) into two terms: 3√x and - x√x.
1. Term 1: 3√x
To find the derivative of √x, we can use the power rule for differentiation. The power rule states that if we have a function f(x) = x^n, then its derivative is f'(x) = nx^(n-1). In this case, n = 1/2 (because the square root √x can be written as x^(1/2)).
Using the power rule, the derivative of √x is (√x)' = (1/2)x^(-1/2).
Now, multiply this derivative by the coefficient 3:
3 * (√x)' = 3 * (1/2)x^(-1/2) = (3/2)x^(-1/2).
Now let's move on to the second term.
2. Term 2: - x√x
To differentiate this term, we need to apply the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product uv is given by (uv)' = u'v + uv'.
Let's assign u(x) = -x and v(x) = √x. Taking the derivatives of u(x) and v(x), we have u'(x) = -1 and v'(x) = (1/2)x^(-1/2).
Now we can apply the product rule:
(-x√x)' = u'v + uv' = (-1) * √x + (-x) * (1/2)x^(-1/2) = -√x - (1/2)x^(1/2).
Now that we have the derivatives of the two terms:
Term 1: 3 * (√x)' = (3/2)x^(-1/2)
Term 2: (-x√x)' = -√x - (1/2)x^(1/2)
The derivative of h(x) = 3√x - x√x is obtained by adding the derivatives of the two terms:
h'(x) = (3/2)x^(-1/2) + (-√x - (1/2)x^(1/2))
Simplifying the expression, we get:
h'(x) = (3/2)x^(-1/2) - √x - (1/2)x^(1/2)
Therefore, the derivative of h(x) = 3√x - x√x is (3/2)x^(-1/2) - √x - (1/2)x^(1/2).