Problem 14 - Thomas Calculus

Question

Find the first derivatives of the functions in Exercises \(11-18\) . $$ z=\cos \left[(1-6 t)^{2 / 3}\right] $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dz}{dt} = 4\sin((1-6t)^{2/3})(1-6t)^{-1/3} \).

Step 1: Identify the Outer Function

The given function is \[ z = \cos \left[(1-6t)^{2/3}\right] \]Here, the outer function is the cosine function, denoted as \( \cos(u) \) where \( u = (1-6t)^{2/3} \). To find \( \frac{dz}{dt} \), we will use the chain rule to differentiate the outer and inner functions.

Step 2: Differentiate the Outer Function Using Chain Rule

To apply the chain rule, first differentiate the outer function \( \cos(u) \) with respect to \( u \):\[ \frac{d}{du} \cos(u) = -\sin(u) \]Thus, using the chain rule, the derivative of \( z \) with respect to \( t \) is:\[ \frac{dz}{dt} = \frac{d}{du} \cos(u) \cdot \frac{du}{dt} = -\sin(u) \cdot \frac{du}{dt} \]

Step 3: Differentiate the Inner Function

The inner function is \( u = (1-6t)^{2/3} \). Differentiate it with respect to \( t \) using the power rule:\[ \frac{du}{dt} = \frac{2}{3}(1-6t)^{-1/3} \cdot (-6) \]Simplifying the derivative of the inner function, we get:\[ \frac{du}{dt} = -4 (1-6t)^{-1/3} \]

Step 4: Combine the Derivatives

Substitute \( \frac{du}{dt} \) back into the expression for \( \frac{dz}{dt} \):\[ \frac{dz}{dt} = -\sin((1-6t)^{2/3}) \cdot -4(1-6t)^{-1/3} \]Simplifying:\[ \frac{dz}{dt} = 4\sin((1-6t)^{2/3})(1-6t)^{-1/3} \]

Step 5: Write the Final Answer

Thus, the derivative of \( z \) with respect to \( t \) can be expressed as:\[ \frac{dz}{dt} = 4\sin((1-6t)^{2/3})(1-6t)^{-1/3} \]

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus for differentiating composite functions. It allows us to take the derivative of a function that is nested within another function. Think of it as peeling away at layers. Each layer requires taking its own derivative before digging deeper.

To use the Chain Rule effectively, you identify two parts: the outer function and the inner function. In our exercise, the outer function is \(\cos(u)\) and the inner function is \((1-6t)^{2/3}\).

To apply the Chain Rule, we first find the derivative of the outer function with respect to the inner variable \(u\). This gives us \(-\sin(u)\). Next, we multiply it by the derivative of the inner function \(u\) with respect to \(t\). This cascade effect is what makes the Chain Rule so powerful. Remember:

• Differentiate the outer function.
• Multiply by the derivative of the inner function.

By following this approach, you will be able to unravel even the most complex functions with ease.

Trigonometric Differentiation

Trigonometric Differentiation involves finding the derivatives of trigonometric functions like sine, cosine, and tangent. These derivatives follow specific rules, which are quite simple once you familiarize yourself with them.

In this problem, we are focusing on the trigonometric function cosine. The derivative of \(\cos(x)\) with respect to \(x\) is \(-\sin(x)\). This negative sign is important because it gives trigonometric functions unique behaviors during differentiation. Knowing these basic derivatives:

• \(\frac{d}{dx} \sin(x) = \cos(x)\)
• \(\frac{d}{dx} \cos(x) = -\sin(x)\)
• \(\frac{d}{dx} \tan(x) = \sec^2(x)\)

Aids tremendously when tackling different calculus problems. In our exercise, we used this rule to simplify the differentiation of the outer function \(\cos(u)\). By applying these known solutions, you can quickly and easily find derivatives of trigonometric expressions.

Power Rule

The Power Rule is another essential differentiation tool used for finding the derivatives of power functions. It's particularly useful when the function you are dealing with is in the form of a power of a variable.

The Power Rule states that if you have a function \(x^n\), its derivative will be \(nx^{n-1}\). It's a straightforward process, involving two simple steps:

• Bring down the exponent as a coefficient.
• Subtract one from the original exponent.

In our exercise, the inner function \(u = (1-6t)^{2/3}\) required the Power Rule.

We had to differentiate with respect to \(t\). This gave us \(-(4)(1-6t)^{-1/3}\). We then integrated this result back into our derivative calculation to apply the Chain Rule. The Power Rule simplifies differentiation significantly by providing a clear, systematic approach to handling powers.