Question:
The derivative of $f(\sec x)$ with respect to $g(\tan x)$ at $x=\frac{\pi}{4}$, where $f^{\prime}(\sqrt{2})=4$ and $g^{\prime}(1)=2$, is
The derivative of $f(\sec x)$ with respect to $g(\tan x)$ at $x=\frac{\pi}{4}$, where $f^{\prime}(\sqrt{2})=4$ and $g^{\prime}(1)=2$, is
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Logic Tip: Notice how the inputs for $f'$ and $g'$ in the given information precisely match the values of $\sec(\pi/4)$ and $\tan(\pi/4)$. If they didn't match, you wouldn't be able to solve the problem! This is an excellent self-checking mechanism during an exam to confirm you are evaluating the chain rule correctly.
Updated On: Apr 28, 2026
- 2
- $\frac{1}{\sqrt{2$
- $\sqrt{2}$
- $\frac{1}{2\sqrt{2$
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The Correct Option is C
Solution and Explanation
Concept:
This requires differentiating a function with respect to another function. Let $u = f(x)$ and $v = g(x)$. The derivative of $u$ with respect to $v$ is given by the parametric differentiation formula: $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. Both $du/dx$ and $dv/dx$ must be evaluated using the Chain Rule.
Step 1: Define the variables and find their derivatives with respect to x.
Let $u = f(\sec x)$ and $v = g(\tan x)$. Differentiate $u$ with respect to $x$ using the chain rule: $$\frac{du}{dx} = f'(\sec x) \cdot \frac{d}{dx}(\sec x)$$ $$\frac{du}{dx} = f'(\sec x) \cdot (\sec x \tan x)$$ Differentiate $v$ with respect to $x$ using the chain rule: $$\frac{dv}{dx} = g'(\tan x) \cdot \frac{d}{dx}(\tan x)$$ $$\frac{dv}{dx} = g'(\tan x) \cdot (\sec^2 x)$$
Step 2: Set up the ratio du/dv and simplify.
$$\frac{du}{dv} = \frac{\frac{du}{dx{\frac{dv}{dx = \frac{f'(\sec x) \cdot \sec x \tan x}{g'(\tan x) \cdot \sec^2 x}$$ Cancel the common factor of $\sec x$: $$\frac{du}{dv} = \frac{f'(\sec x) \cdot \tan x}{g'(\tan x) \cdot \sec x}$$
Step 3: Evaluate the derivative at the specified point.
We need to find the value at $x = \frac{\pi}{4}$. First, calculate the required trigonometric values: $$\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$$ $$\tan\left(\frac{\pi}{4}\right) = 1$$ Substitute these into the simplified ratio: $$\left. \frac{du}{dv} \right|_{x=\frac{\pi}{4 = \frac{f'(\sqrt{2}) \cdot (1)}{g'(1) \cdot (\sqrt{2})}$$ We are given that $f'(\sqrt{2}) = 4$ and $g'(1) = 2$. Substitute these given values: $$\left. \frac{du}{dv} \right|_{x=\frac{\pi}{4 = \frac{4 \cdot 1}{2 \cdot \sqrt{2 = \frac{4}{2\sqrt{2 = \frac{2}{\sqrt{2$$ Rationalize the expression: $$\frac{2}{\sqrt{2 = \sqrt{2}$$
This requires differentiating a function with respect to another function. Let $u = f(x)$ and $v = g(x)$. The derivative of $u$ with respect to $v$ is given by the parametric differentiation formula: $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. Both $du/dx$ and $dv/dx$ must be evaluated using the Chain Rule.
Step 1: Define the variables and find their derivatives with respect to x.
Let $u = f(\sec x)$ and $v = g(\tan x)$. Differentiate $u$ with respect to $x$ using the chain rule: $$\frac{du}{dx} = f'(\sec x) \cdot \frac{d}{dx}(\sec x)$$ $$\frac{du}{dx} = f'(\sec x) \cdot (\sec x \tan x)$$ Differentiate $v$ with respect to $x$ using the chain rule: $$\frac{dv}{dx} = g'(\tan x) \cdot \frac{d}{dx}(\tan x)$$ $$\frac{dv}{dx} = g'(\tan x) \cdot (\sec^2 x)$$
Step 2: Set up the ratio du/dv and simplify.
$$\frac{du}{dv} = \frac{\frac{du}{dx{\frac{dv}{dx = \frac{f'(\sec x) \cdot \sec x \tan x}{g'(\tan x) \cdot \sec^2 x}$$ Cancel the common factor of $\sec x$: $$\frac{du}{dv} = \frac{f'(\sec x) \cdot \tan x}{g'(\tan x) \cdot \sec x}$$
Step 3: Evaluate the derivative at the specified point.
We need to find the value at $x = \frac{\pi}{4}$. First, calculate the required trigonometric values: $$\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$$ $$\tan\left(\frac{\pi}{4}\right) = 1$$ Substitute these into the simplified ratio: $$\left. \frac{du}{dv} \right|_{x=\frac{\pi}{4 = \frac{f'(\sqrt{2}) \cdot (1)}{g'(1) \cdot (\sqrt{2})}$$ We are given that $f'(\sqrt{2}) = 4$ and $g'(1) = 2$. Substitute these given values: $$\left. \frac{du}{dv} \right|_{x=\frac{\pi}{4 = \frac{4 \cdot 1}{2 \cdot \sqrt{2 = \frac{4}{2\sqrt{2 = \frac{2}{\sqrt{2$$ Rationalize the expression: $$\frac{2}{\sqrt{2 = \sqrt{2}$$
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