Question

Suppose you forgot the Quotient Rule for calculating \(\frac{d}{d x}\left(\frac{f(x)}{g(x)}\right) .\) Use the Chain Rule and Product Rule with the identity \(\frac{f(x)}{g(x)}=f(x)(g(x))^{-1}\) to derive the Quotient Rule.

Short answer

Question: Derive the Quotient Rule using the Chain Rule, Product Rule, and the identity \(\frac{f(x)}{g(x)} = f(x)(g(x))^{-1}\). Answer: The Quotient Rule is given by \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\).

Step-by-step solution

Rewrite the expression using the given identity

Using the given identity, we rewrite \(\frac{f(x)}{g(x)}\) as \(f(x)(g(x))^{-1}\).

Apply the Product Rule

The Product Rule states that \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\) for functions \(u\) and \(v\). In our case, \(u = f(x)\) and \(v = (g(x))^{-1}\). So, we get: \(\frac{d}{dx}\left(f(x)(g(x))^{-1}\right) = f(x)\frac{d((g(x))^{-1})}{dx} + (g(x))^{-1}\frac{df(x)}{dx}\)

Apply the Chain Rule to \(\frac{d}{dx}((g(x))^{-1})\)

The Chain Rule states that \(\frac{d}{dx}(h(g(x))) = h'(g(x))\frac{d}{dx}g(x)\) for functions \(h\) and \(g\). Here, \(h(u) = u^{-1}\) and \(g(x) = g(x)\). So, we get: \(\frac{d}{dx}((g(x))^{-1}) = \left(\frac{d}{du}(u^{-1})\right)_{u=g(x)} \cdot \frac{d}{dx}(g(x)) = - (g(x))^{-2} \cdot g'(x)\)

Replace \(\frac{d}{dx}((g(x))^{-1})\) in the expression from Step 2

Now, we can replace \(\frac{d}{dx}((g(x))^{-1})\) in the expression found in Step 2: \(\frac{d}{dx}\left(f(x)(g(x))^{-1}\right) = f(x)\left(- (g(x))^{-2} \cdot g'(x)\right) + (g(x))^{-1}\frac{df(x)}{dx}\)

Simplify and write out the Quotient Rule

Finally, we simplify the expression: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = -\frac{f(x)g'(x)}{g(x)^2} + \frac{f'(x)}{g(x)}\) Now we can rewrite it in the usual form of the Quotient Rule: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\) This is the Quotient Rule derived using the Chain Rule and Product Rule.