Question

Derivatives Find and simplify the derivative of the following functions. $$y=(3 t-1)(2 t-2)^{-1}$$

Short answer

The derivative of the function is: $$y' = \frac{3}{2t - 2} - \frac{6(3t - 1)}{(2t - 2)^2}$$

Step-by-step solution

Identify the functions

Identify the functions in the product: \(u = 3t - 1\) and \(v = (2t - 2)^{-1}\).

Find the derivative of the first function

Differentiate the first function \(u\) with respect to \(t\): $$u' = \frac{d}{dt} (3t - 1) = 3$$

Apply the chain rule to find the derivative of the second function

We can treat \((2t - 2)^{-1}\) as \(h(g(t))\), where \(h(u) = u^{-1}\) and \(g(t) = 2t - 2\). Find the derivatives of \(h(u)\) and \(g(t)\): $$h'(u) = \frac{d}{du}(u^{-1}) = -u^{-2}$$ $$g'(t) = \frac{d}{dt}(2t - 2) = 2$$ Now apply the chain rule: $$v'(t) = h'(g(t)) \cdot g'(t) = - (2t - 2)^{-2} \cdot 2$$

Apply the product rule

Now we will apply the product rule to find the derivative of the entire function \(y\): $$y' = u'v + uv' = 3 (2t - 2)^{-1} + (3t - 1)(- (2t - 2)^{-2} \cdot 2)$$

Simplify the derivative

Now, we will simplify the expression: $$y' = 3 (2t - 2)^{-1} - 2(3t - 1)(2t - 2)^{-2}$$ The derivative of the given function is: $$y' = \frac{3}{2t - 2} - \frac{6(3t - 1)}{(2t - 2)^2}$$

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function is essentially a function within another, defined as \( h(g(t)) \). For instance, in our original problem,

• \( h(u) = u^{-1} \)
• \( g(t) = 2t - 2 \)

To differentiate a composite function, you first differentiate the outer function \( h \) with respect to its inner function \( u \), and then multiply by the derivative of the inner function \( g(t) \). This leads us to the chain rule formula:\[v'(t) = h'(g(t)) \cdot g'(t)\] Remember, the key to applying the chain rule is identifying the inner and outer functions correctly. In practice, finding each derivative step-by-step and then multiplying them together will yield the desired result.
Understanding the chain rule is essential when dealing with nested functions in calculus.

Applying the Product Rule

When you have two functions multiplied together, and you need to find the derivative, the product rule comes into play. This rule helps us differentiate expressions where each component is a separate function.
For our exercise:

• \( u = 3t - 1 \)
• \( v = (2t - 2)^{-1} \)

The product rule formula is given by:\[y' = u'v + uv'\]This formula simply states that to take the derivative of the product of two functions, you need to:

• Differentiate the first function \( (u') \)
• Multiply by the second function unchanged \( (v) \)
• Add to this the first function unchanged \( (u) \)
• Multiplied by the derivative of the second function \( (v') \)

Using this technique ensures we correctly find the derivative of products of functions, making it a crucial tool in our calculus toolbox.

Basics of Differentiation

Differentiation is one of the core processes in calculus and involves finding the derivative of a function. The derivative shows how a function changes at any given point and is essentially the slope of the function at that point.
Here are several foundational principles that guide differentiation:

• The derivative of a constant is zero.
• The derivative of \( t^n \) is \( nt^{n-1} \), where \( n \) is a real number.
• Summation: The derivative of a sum is simply the sum of the derivatives.

In practice, combining the chain and product rules often becomes necessary, as seen in our exercise. Differentiation manifests in many real-world applications, from calculating velocities to predicting trends. Grasping these basic principles paves the way for deeper understanding and application of calculus in various fields.