Math Problem Statement

derivative

Find the derivative of the function u = ∛(t²) + 2√(t³).

Solution

To find the derivative of the function given in the image:

u=t23+2t3u = \sqrt[3]{t^2} + 2\sqrt{t^3}

We need to apply the appropriate rules of differentiation to each term individually. Let's calculate the derivatives of each term.

  1. For the first term, t23\sqrt[3]{t^2}, we can rewrite it as t2/3t^{2/3}. Using the power rule for differentiation, ddttn=ntn1\frac{d}{dt} t^n = n t^{n-1}, the derivative is: ddtt2/3=23t2/31=23t1/3\frac{d}{dt} t^{2/3} = \frac{2}{3} t^{2/3 - 1} = \frac{2}{3} t^{-1/3}

  2. For the second term, 2t32\sqrt{t^3}, we can rewrite it as 2t3/22t^{3/2}. Using the power rule again, the derivative is: ddt2t3/2=232t3/21=3t1/2\frac{d}{dt} 2t^{3/2} = 2 \cdot \frac{3}{2} t^{3/2 - 1} = 3t^{1/2}

Therefore, combining these derivatives, the derivative uu' of the function uu is: u=23t1/3+3t1/2u' = \frac{2}{3} t^{-1/3} + 3t^{1/2}

Which can be written as: u=231t3+3tu' = \frac{2}{3} \frac{1}{\sqrt[3]{t}} + 3 \sqrt{t}

If you have any questions or need further details on any of the steps, feel free to ask! Here are some follow-up topics you might consider:

  1. How to simplify the expression of the derivative further if possible.
  2. How to evaluate the derivative at a specific value of tt.
  3. The importance of understanding different forms of exponents when differentiating.
  4. How to graph the original function and its derivative to visualize their behaviors.
  5. How the behavior of the derivative reflects on the increasing or decreasing nature of the original function.

Tip: When differentiating expressions involving roots and powers, it’s useful to first rewrite them in exponential form using rational exponents to simplify differentiation using the power rule.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Power Rule
Rational Exponents

Formulas

d/dt [t^n] = n * t^(n-1)
u = ∛(t²) + 2√(t³) can be rewritten as u = t^(2/3) + 2t^(3/2)

Theorems

Power Rule of Differentiation

Suitable Grade Level

Grades 11-12 (Calculus I)