Math Problem Statement

Find the derivative of the function q = sin(t / sqrt(t + 3)).

Solution

We are tasked with finding the derivative of the function:

q=sin(tt+3)q = \sin\left( \frac{t}{\sqrt{t + 3}} \right)

We will use the chain rule to differentiate this function.

Step-by-step solution:

  1. Outer function: The outer function is sin(u)\sin(u), where u=tt+3u = \frac{t}{\sqrt{t + 3}}. The derivative of sin(u)\sin(u) with respect to uu is:

ddusin(u)=cos(u)\frac{d}{du} \sin(u) = \cos(u)

Thus, applying the chain rule:

dqdt=cos(tt+3)dudt\frac{dq}{dt} = \cos\left( \frac{t}{\sqrt{t + 3}} \right) \cdot \frac{du}{dt}

  1. Inner function: Now we need to find dudt\frac{du}{dt} where u=tt+3u = \frac{t}{\sqrt{t + 3}}.

    • Use the quotient rule for the derivative of u=tt+3u = \frac{t}{\sqrt{t + 3}}.
    • Let f(t)=tf(t) = t and g(t)=t+3g(t) = \sqrt{t + 3}. The quotient rule states:

ddt(f(t)g(t))=f(t)g(t)f(t)g(t)[g(t)]2\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}

  • f(t)=1f'(t) = 1 (the derivative of tt),
  • g(t)=t+3g(t) = \sqrt{t + 3}, so g(t)=12t+3g'(t) = \frac{1}{2\sqrt{t + 3}} (the derivative of a square root function).

Now applying the quotient rule:

dudt=1t+3t12t+3(t+3)2\frac{du}{dt} = \frac{1 \cdot \sqrt{t + 3} - t \cdot \frac{1}{2\sqrt{t + 3}}}{(\sqrt{t + 3})^2}

  1. Simplifying the expression:

The denominator simplifies to t+3t + 3, and the numerator becomes:

t+31t2t+3=2(t+3)t2t+3=t+62t+3\frac{\sqrt{t + 3}}{1} - \frac{t}{2\sqrt{t + 3}} = \frac{2(t + 3) - t}{2\sqrt{t + 3}} = \frac{t + 6}{2\sqrt{t + 3}}

Thus, the derivative dudt\frac{du}{dt} simplifies to:

dudt=t+62(t+3)3/2\frac{du}{dt} = \frac{t + 6}{2(t + 3)^{3/2}}

  1. Final derivative: Now substitute back into the original expression for dqdt\frac{dq}{dt}:

dqdt=cos(tt+3)t+62(t+3)3/2\frac{dq}{dt} = \cos\left( \frac{t}{\sqrt{t + 3}} \right) \cdot \frac{t + 6}{2(t + 3)^{3/2}}

Answer:

dqdt=cos(tt+3)t+62(t+3)3/2\frac{dq}{dt} = \cos\left( \frac{t}{\sqrt{t + 3}} \right) \cdot \frac{t + 6}{2(t + 3)^{3/2}}

Let me know if you want further details or explanations.

Here are 5 related questions for deeper understanding:

  1. How does the chain rule apply in this context, and why is it necessary?
  2. What other functions, besides sine, commonly require the chain rule in their derivatives?
  3. How would the derivative change if the function involved a cosine instead of a sine?
  4. Can you derive the derivative of sin(1t)\sin\left(\frac{1}{t}\right) and compare the process?
  5. How does the quotient rule simplify complex fractions in calculus?

Tip: The quotient rule is especially useful when differentiating expressions involving two functions divided by each other, but it’s important to apply it carefully to avoid algebraic errors.

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Math Problem Analysis

Mathematical Concepts

Calculus
Chain Rule
Quotient Rule
Trigonometry

Formulas

Chain Rule: d/dt[f(g(t))] = f'(g(t)) * g'(t)
Quotient Rule: d/dt[f(t)/g(t)] = (f'(t)g(t) - f(t)g'(t)) / [g(t)]^2
Derivative of sin(u) = cos(u)

Theorems

Chain Rule
Quotient Rule

Suitable Grade Level

Grades 11-12 / Early College