Problem 50 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ g(t)=\left(\frac{1}{\sin t^{2}}\right)^{3 / 2} $$

Step-by-Step Solution

Answer

The derivative is \(-3t \cdot \left(\sin(t^2)\right)^{-5/2} \cdot \cos(t^2)\)."

Step 1: Expression Reformatting

First, we rewrite the function into a more familiar form. The given function is \( g(t) = \left( \frac{1}{\sin(t^2)} \right)^{3/2} \). We recognize this as \( g(t) = \left( \sin(t^2)\right)^{-3/2} \). This will help us use power rule and chain rule to take the derivative.

Step 2: Differentiate Using the Power Rule and Chain Rule

We apply the power rule and chain rule. The power rule says \( (x^n)' = n \cdot x^{n-1} \), where in this case, \( x = \sin(t^2) \) and \( n = -\frac{3}{2} \). We have:\[ \frac{d}{dt} \left(\sin(t^2)\right)^{-3/2} = -\frac{3}{2} \left(\sin(t^2)\right)^{-5/2} \cdot \frac{d}{dt} (\sin(t^2)) \]

Step 3: Differentiate Inner Function

Now we find \( \frac{d}{dt} (\sin(t^2)) \) using the chain rule. Chain rule states if \( y = u(v(t)) \), then \( \frac{dy}{dt} = \frac{dy}{dv} \cdot \frac{dv}{dt} \). Here, \( u = \sin(u) \) and \( v = t^2 \).\[ \frac{d}{dt} \sin(t^2) = \cos(t^2) \cdot \frac{d}{dt}(t^2) = \cos(t^2) \cdot 2t \]

Step 4: Combine Results

Combine the results from Step 2 and Step 3. Substitute \( \frac{d}{dt} (\sin(t^2)) = \cos(t^2) \cdot 2t \) back into the expression from Step 2:\[ -\frac{3}{2} \left(\sin(t^2)\right)^{-5/2} \cdot (\cos(t^2) \cdot 2t) \]

Step 5: Simplify the Expression

Simplify the expression:\[ -3t \cdot \left(\sin(t^2)\right)^{-5/2} \cdot \cos(t^2) \]This gives us the derivative of the original function.

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used to find the derivative of composite functions. When you encounter a function within another function, the chain rule comes into play. It allows us to differentiate these complex layers one at a time.
For a function defined as the composition of two functions, expressed mathematically as \( y = u(v(t)) \), the chain rule states that the derivative \( \frac{dy}{dt} \) is equal to the product of the derivative of \( u \) with respect to \( v \) (notated as \( \frac{du}{dv} \)) and the derivative of \( v \) with respect to \( t \) (notated as \( \frac{dv}{dt} \)).
- **Break Down**: \( u = \sin(u) \), \( v = t^2 \).- **Derivatives**: Find \( \frac{du}{dv} \) which is \( \cos(v) \), and \( \frac{dv}{dt} \) which is \( 2t \).By multiplying the two derivatives together: \( \frac{d}{dt} \sin(t^2) = \cos(t^2) \cdot 2t \). This is how we apply the chain rule.

Power Rule

The power rule provides a straightforward way to differentiate functions of the form \( x^n \). When we need to find the derivative of a term like this, the power rule comes into play.
This rule is described mathematically as \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \). It tells us to multiply the original power by the coefficient in front and then subtract one from the power.- **Our Application**: The function is \( \left(\sin(t^2)\right)^{-3/2} \), which is a composite function with an outer power of \(-3/2\).- **Using the Rule**: The derivative using power rule gives \( -\frac{3}{2} \cdot \left(\sin(t^2)\right)^{-5/2} \).By applying the power rule, we differentiated the outermost layer while acknowledging the inner function, readying it for the chain rule.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental components in calculus. They have distinctive behaviors and derivatives that make them unique.
- **Key Functions**: The sine function \( \sin(x) \) and the cosine function \( \cos(x) \).Each of these functions has a standard derivative:- \( \frac{d}{dx}\sin(x) = \cos(x) \)- \( \frac{d}{dx}\cos(x) = -\sin(x) \)In the original exercise, we needed to differentiate \( \sin(t^2) \). By recognizing that the derivative of \( \sin(x) \) is \( \cos(x) \), and then applying the chain rule, we successfully found the derivative \( \cos(t^2) \cdot 2t \).
Understanding these properties makes handling trigonometric functions in calculus more approachable.