Question

\(\mathrm{f}\) and \(\mathrm{g}\) are two differentiable function and fog \(=\mathrm{I}\) (Identify function) and \(g^{\prime}(a)=2, g(a)=b\) then \(f^{\prime}(b)=\) (a) 2 (b) \((1 / 2)\) (c) \(-2\) (d) \(-(1 / 2)\)

Short answer

The short answer is: \(f'(b) = \frac{1}{2}\).

Step-by-step solution

Differentiate the given equation

Differentiate both sides of the equation f(g(x)) = x. Since \(\mathrm{f}\) and \(\mathrm{g}\) are differentiable, we can use the chain rule to differentiate \[f(g(x))\] with respect to x. The chain rule states: \[\frac{\mathrm{d}(f(g(x)))}{\mathrm{d}x} = f ' (g(x))\cdot g'(x)\] So, let's apply the chain rule to our provided equation \(f(g(x)) = x\) and find the derivative on both sides. \[\frac{\mathrm{d}(f(g(x)))}{\mathrm{d}x} = \frac{\mathrm{d}x}{\mathrm{d}x}\] \[f'(g(x))\cdot g'(x) = 1\] Now we have an equality involving the derivatives of f and g.

Substitute the given values

We are given that g'(a) = 2 and g(a) = b. We want to find f'(b). To do this, let's substitute the values into the equation derived in step 1. Let \(x = a\): \[f'(g(a))\cdot g'(a) = 1\] Now, substituting the provided values: \[f'(b)\cdot 2 = 1\]

Solve for f'(b)

In the previous step, we have the equation: \[f'(b)\cdot 2 = 1\] Now let's solve it for \(f'(b)\): \[f'(b) = \frac{1}{2}\] So, the correct answer is (b) \(\frac{1}{2}\).

Key concepts

Function Composition

Understanding function composition is like understanding a recipe where one ingredient is itself a combination of other ingredients. Imagine a scenario where you take a fruit, blend it into a smoothie (function g), and then freeze it to make a popsicle (function f). In mathematics, the function composition of f and g, denoted as f(g(x)), is like making a popsicle from the smoothie, where the input x represents the original fruit.

In the context of our exercise, think of g as a function that transforms an input x into an output g(x), and f as another function that takes this output g(x) and converts it to f(g(x)). When the composition of both functions results in the original input x, we say that f is the inverse of g. This inverse relationship is crucial when solving problems involving the derivative of composed functions.

Derivative of Inverse Functions

The derivative of inverse functions can sometimes be a mind-bending concept, but it's less mystifying when we realize that it's all about flipping a function on its side. Picture a function like a winding road, where the derivative provides the slope at any point. Now, imagine reversing that journey, effectively driving on the inverse road. The slope at any point on this return trip is the derivative of the inverse function.

In our problem, f and g are like two roads that reverse each other's journey – one unwinding what the other winds. When we differentiate the identity function, which is like a straight road that doesn't change direction, we get a derivative that tells us how f and g interact at their respective points. For g(a) = b, it's like saying we're looking at the same spot on the road, but from the perspective of the return trip. Deriving the value for f'(b) is akin to finding the slope (rate of change) of the winding road for the reverse journey at that particular spot.

Chain Rule Application

When you put on a chain, you link several individual pieces together to form a connected whole. The chain rule in calculus serves a similar purpose: it connects the derivatives of functions when one function is composed of another. This rule is a powerful tool for differentiating complex expressions, where one function is nestled inside another, like a set of Russian dolls.

In our exercise, we applied the chain rule to find the derivative of the composed function f(g(x)). By taking the derivative of the outer function f at g(x), and then multiplying it by the derivative of the inner function g at x, we get the derivative of the overall composition with respect to x. This multi-step method is necessary to unravel the layered changes occurring within function compositions and is a critical concept in calculus that allows us to handle our mathematical ingredients with precision, just like a skilled chef.