Is this correct take the derivative of

• laker_gurl3
In summary, the conversation involved a discussion about the correct answer to a product rule problem. One person provided their answer of (t-1)^-1/2 t^-3 {-3/2t +2}, while the other person corrected it to the form \frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right] t^{-3} \left(4-3t\right). The first person then asked for clarification on a different problem, but the second person was unsure of what they were asking. The conversation ended with the second person explaining how they got their corrected answer.
laker_gurl3
(t-1)^1/2*(t^-2)

I hope you guys can understand what I am trying to say up there... SO i did the product rule, and my answer was this.. lemmi know if it's correct...thanks a bunch.

(t-1)^-1/2 t^-3 { -3/2t +2 }

I get

$$\frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right] t^{-3} \left(4-3t\right)$$

Which is the same thing,so everything is okay.

Daniel.

P.S.Lakers missed the play-offs

Last edited:
hey that looks perfect to me!

is this what you trying to differentiate - $$(t-1)^{\frac{t^2}{2}}$$ or is this
$$(t-1)^{\frac{1}{2}} \frac{t^2}{2}$$

Nope.

$$(t-1)^{\frac{1}{2}} t^{-2}$$

Daniel.

dextercioby said:
Nope.

$$(t-1)^{\frac{1}{2}} t^{-2}$$

Daniel.
I can see where the $$\frac{1}{2}\left[(t-1)^{-\frac{1}{2}}\right]$$ came from but the $$t^{-3} \left(4-3t\right)$$ has lost me. What did you do to get that because I would have just done $$-2t^{-3}$$

I forced something as a factor (v.above),and that's how i ended up with that paranthesis.

Daniel.

1. What is a derivative and why is it important?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is important because it helps us understand how a function changes over time, and is crucial in many areas of science, such as physics and economics.

2. How do you take the derivative of a function?

To take the derivative of a function, you use a set of rules and formulas known as calculus. The specific method for taking the derivative depends on the type of function, but it generally involves finding the slope of the function at a specific point.

3. Is there a shortcut or simpler way to take the derivative?

Yes, there are various shortcuts and techniques that can be used to simplify the process of taking a derivative. These include the power rule, product rule, quotient rule, and chain rule. It is important to understand these rules in order to efficiently take derivatives of more complex functions.

4. Can you explain the application of derivatives in real-world scenarios?

Derivatives have numerous real-world applications. For example, in physics, derivatives are used to calculate velocity and acceleration. In economics, derivatives are used to analyze and predict changes in supply and demand. In engineering, derivatives are used to optimize designs and improve efficiency.

5. What are some common mistakes when taking derivatives?

Some common mistakes when taking derivatives include forgetting to apply the chain rule, mixing up signs or exponents, and not simplifying the final answer. It is important to double-check your work and use the appropriate rules and formulas to avoid these mistakes.

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