SOLUTION: If f(x)=1/³√2x+1 find f(x+h)-f(x)/h at h=0

Question 1190765: If f(x)=1/³√2x+1 find f(x+h)-f(x)/h at h=0
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
You're asking for the derivative of f(x) with respect to x, which is the limit of [f(x+h) - f(x)] / h as h approaches 0. Since you're asking to evaluate it *at* h=0, you're effectively asking for the derivative.
Here's how to find f'(x):
1. **Rewrite f(x):** It's often easier to work with fractional exponents: f(x) = (2x + 1)^(-1/3)
2. **Apply the chain rule:**
* The outer function is u^(-1/3), where u = 2x + 1. The derivative of this is (-1/3)u^(-4/3).
* The inner function is u = 2x + 1. The derivative of this is 2.
3. **Combine the derivatives:**
f'(x) = (-1/3)(2x + 1)^(-4/3) * 2
f'(x) = (-2/3)(2x + 1)^(-4/3)
4. **Rewrite with a radical (optional):**
f'(x) = -2 / (3 * ³√((2x + 1)^4)) or f'(x) = -2 / (3(2x+1) * ³√(2x+1))
Since the question asks to evaluate at h=0, but the expression contains only x, we presume that it is asking for the derivative of f(x) with respect to x. Therefore, the answer is:
f'(x) = (-2/3)(2x + 1)^(-4/3) or f'(x) = -2 / (3(2x+1) * ³√(2x+1))

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