Derivative of s/sqrt(60s + 25) | Calculus Homework Solution

  • Thread starter Appleton
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The general formula for that is d/dx(f(x) g(x)) = f(x) g'(x) + f'(x) g(x).In this case, f(x) = s and g(x) = (60s + 25)-1/2.In summary, when using the chain rule to calculate the derivative of s/sqrt(60s + 25), we need to rewrite it as s(60s + 25)^-1/2 and use the formula d/dx(f(x) g(x)) = f(x) g'(x) + f'(x) g(x). This gives us a final answer of -30s(60s + 25)^-3/2.
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Appleton
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Homework Statement



d/ds s/sqrt(60s + 25)

Homework Equations





The Attempt at a Solution



d/ds s/sqrt(60s + 25) = d/ds s(60s + 25)^-1/2
if a = s(60s + 25)^-1/2
and b = (60s + 25)
da/db = -1/2s(60s +25)^-3/2
db/ds = 60
da/ds = -30s(60s +25)^-3/2

Any help much appreciated.
 
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  • #2
It looks like you are trying to use the chain rule, but that doesn't quite work here. The reason is that if you want to calculate da/db, you have to write a completely in terms of b, i.e. as (b - 25) / 60 * b-1/2.

Note that after the rewriting in your first line (s (60s + 25)-1/2) you have a product of two functions, i.e. you are looking for d/ds(a b) with a = s, b = (60s + 25)-1/2
 

1. What does the equation D/ds s/sqrt(60s + 25) represent?

The equation represents the derivative of s divided by the square root of 60s + 25.

2. What is the purpose of using the square root in the equation?

The square root helps to find the rate of change of the function with respect to s.

3. How is the derivative of the equation calculated?

The derivative is calculated using the quotient rule of differentiation, which states that the derivative of s/sqrt(60s + 25) is (sqrt(60s + 25) - s(60s + 25)^(-1/2)) / (60s + 25).

4. What are the possible applications of this equation in scientific research?

This equation can be used to analyze and understand the relationship between two variables in a system, such as the rate of change of a biological process over time.

5. Can this equation be simplified further?

Yes, this equation can be simplified by using algebraic manipulation and simplifying the square root term. However, the resulting equation may not be as informative or useful for scientific analysis.

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