- #1
ttpp1124
- 110
- 4
- Homework Statement
- I'm supposed to differentiate, but I was asked not to simplify..I'm not sure if my final answer is formatted such that it can be understood...
- Relevant Equations
- n/a
Differentiating y = (x -1)^3 (5√2x^2 -1)
- Thread starter ttpp1124
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- Differentiating
In summary, the process of differentiating this function involves using the power rule and the product rule. The derivative of the function is 3(x-1)^2(5√2x^2-1) + 20√2x^2(x-1)^3. The function can be simplified further by expanding and combining like terms. The domain is all real numbers and the range is also all real numbers.
- #2
Mark44
Mentor
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- 9,862
Looks OK, but the second term is really hard to fathom.
If you're using Word or something similar, your exponents like (4/5) need to have braces -- { } -- around them.
In LaTeX, and cleaned up a bit by making both expressions involve exponents, and getting rid of unneeded parentheses, your answer looks like this:
$$\frac{dy}{dx} = 3(x - 1)^2 (2x^2 - 1)^{1/5} + \frac{4x(x - 1)^3}{5(2x^2 - 1)^{4/5}} $$
If you're using Word or something similar, your exponents like (4/5) need to have braces -- { } -- around them.
In LaTeX, and cleaned up a bit by making both expressions involve exponents, and getting rid of unneeded parentheses, your answer looks like this:
$$\frac{dy}{dx} = 3(x - 1)^2 (2x^2 - 1)^{1/5} + \frac{4x(x - 1)^3}{5(2x^2 - 1)^{4/5}} $$
1. What is the derivative of y = (x -1)^3 (5√2x^2 -1)?
The derivative of y = (x -1)^3 (5√2x^2 -1) is given by the product rule:
y' = (x -1)^3 * d/dx(5√2x^2 -1) + (5√2x^2 -1) * d/dx(x -1)^3
Applying the power rule and chain rule, the derivative simplifies to:
y' = 3(x -1)^2 * 5√2x + (x -1)^3 * 5√2x / √2x
Simplifying further, the derivative is:
y' = 15(x -1)^2 √2x + 5√2x * (x -1)^2
2. How do you find the critical points of y = (x -1)^3 (5√2x^2 -1)?
To find the critical points, we need to take the derivative of the function and set it equal to 0.
y' = 15(x -1)^2 √2x + 5√2x * (x -1)^2 = 0
Simplifying and solving for x, we get two critical points:
x = 1/√2 and x = 1.
To determine if these are maximum or minimum points, we can use the second derivative test.
3. What is the second derivative of y = (x -1)^3 (5√2x^2 -1)?
The second derivative of y = (x -1)^3 (5√2x^2 -1) is given by:
y'' = 30(x -1)√2 + 15(x -1)^2 / √2x + 10√2x(x -1)^2
Simplifying further, we get:
y'' = 30(x -1)√2 + 15(x -1)^2 / √2x + 10√2x(x -1)^2
4. How do you determine the concavity of y = (x -1)^3 (5√2x^2 -1)?
To determine the concavity, we need to look at the sign of the second derivative.
If y'' > 0, the function is concave up.
If y'' < 0, the function is concave down.
In this case, we have y'' = 30(x -1)√2 + 15(x -1)^2 / √2x + 10√2x(x -1)^2.
We can plug in the critical points found earlier (x = 1/√2 and x = 1) to determine the concavity of the function.
5. What is the domain of y = (x -1)^3 (5√2x^2 -1)?
The domain of y = (x -1)^3 (5√2x^2 -1) is all real numbers except for x = 1/√2 and x = 1, as these values would result in division by 0.
Therefore, the domain is:
(-∞, 1/√2) U (1/√2, 1) U (1, ∞)
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