Question

Find the derivative of the inverse of the following finctions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\). $$f(x)=3 x+4 ;(16,4)$$

Short answer

Answer: The derivative of the inverse function at point (16, 4) is (f^(-1))'(4) = 1/3.

Step-by-step solution

Find the derivative of the given function, \(f(x)\)#

Since we are given the function \(f(x)=3x+4\), we can find its derivative, \(f'(x)\) using the power rule. $$f'(x) = \frac{d}{dx}(3x + 4)$$ $$f'(x) = 3$$

Find the value of \(f'(f^{-1}(y))\)#

Now that we have derived f'(x), we can substitute \(f^{-1}(4)\) into it. $$f'(f^{-1}(4)) = f'(16) = 3$$

Calculate the derivative of the inverse function \((f^{-1})'(y)\) using the given formula#

Now we can use the formula to find \((f^{-1})'(y)\): $$(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$$ Substitute the value we found in Step 2: $$(f^{-1})'(4) = \frac{1}{3}$$ Thus, the derivative of the inverse function at the specified point \((16, 4)\) is \((f^{-1})'(4) = \frac{1}{3}\).

Key concepts

Chain Rule

When calculating derivatives of composite functions, the chain rule is an indispensable tool. In essence, it provides a method to decompose the differentiation process of complex functions into simpler parts. For example, consider a function represented as \(h(x) = g(f(x))\). The derivative of this composite function, according to the chain rule, is \(h'(x) = g'(f(x)) \cdot f'(x)\). This means you first find the derivative of the outer function \(g\) with respect to its inner function \(f\), and then multiply it by the derivative of the inner function \(f\) with respect to \(x\).

Applying the chain rule helps in finding derivatives of inverse functions as well. To find the derivative of an inverse function at a particular point, you can use the chain rule alongside the known derivatives. For instance, if you know \(f'(x)\) and you need \(f^{-1})'(y)\), the chain rule can facilitate this calculation. As illustrated in the main exercise, after finding \(f'(x)\), we evaluated it at the inverse function, which is a chain rule application that leads to the identification of \(f^{-1})'(y)\).

Power Rule

The power rule is a basic but powerful technique for differentiating functions of the form \(x^n\) where \(n\) is any real number. The rule states that if you have a function \(f(x) = x^n\), its derivative is \(f'(x) = n \cdot x^{n-1}\). This simplifies the differentiation process because it removes the need for more complex limit definitions of the derivative.

In the context of our problem, the power rule is directly applied to find the derivative of a linear function \(f(x) = 3x + 4\). Since the exponent in \(3x\) is 1 (as we can write \(x\) as \(x^1\)), applying the power rule gives us the derivative as \(3 \cdot x^{1-1} = 3 \cdot x^0 = 3\). The constant term 4 drops out because the derivative of any constant is zero. This illustrates how we use the power rule to quickly find the derivative of the given function without much complication.

Inverse Functions

Inverse functions essentially reverse the effect of a function. If you have a function \(f(x)\), its inverse, denoted \(f^{-1}(x)\), is a function that, when applied to \(f(x)\), returns the original value \(x\). Mathematically, if \(f(a) = b\), then \(f^{-1}(b) = a\).

Finding derivatives of inverse functions can seem daunting because often the inverse function itself is not explicitly known. However, there's a clever workaround using derivatives: The derivative of \(f^{-1}(y)\) at any point \(y\) can be computed as \(\frac{1}{f'(f^{-1}(y))}\). What this expression tells us is that you don't need the formula for \(f^{-1}(x)\) itself; you just need the derivative of \(f(x)\), which you can then evaluate at \(f^{-1}(y)\). This concept not only simplifies the process but also emphasizes the deep connection between a function and its inverse through differentiation. In the provided example, knowing the value of \(f'(x)\) at the point \(16\), we immediately discovered the slope of the tangent to the curve of the inverse function at \(4\), without explicitly finding the form of \(f^{-1}(x)\).