Question:
If \(f\) is a derivable function and
\[
2f(\sin x)+f(\cos x)=x
\qquad \forall x\in \mathbb{R},
\]
then
\[
f'(x)=
\]
If \(f\) is a derivable function and
\[
2f(\sin x)+f(\cos x)=x
\qquad \forall x\in \mathbb{R},
\]
then
\[
f'(x)=
\]
Show Hint
For equations involving both \(f(\sin x)\) and \(f(\cos x)\), differentiate once and then replace
\[
x \mapsto \frac{\pi}{2}-x
\]
to obtain a second equation. Solving the pair usually gives \(f'\).
Updated On: Jul 9, 2026
- \[ \sin x+\cos x \]
- \[ \sin x-\cos x \]
- \[ \sqrt{1-x^2} \]
- \[ \frac{1}{\sqrt{1-x^2}} \] \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Use the given functional equation and differentiate it. Then eliminate the second derivative term by replacing \(x\) with \(\frac{\pi}{2}-x\).
Step 1: Differentiate the given equation. Given \[ 2f(\sin x)+f(\cos x)=x. \] Differentiating w.r.t. \(x\), \[ 2f'(\sin x)\cos x - f'(\cos x)\sin x = 1. \] \[ \cdots (1) \]
Step 2: Replace \(x\) by \(\frac{\pi}{2}-x\). Since \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x, \] equation (1) becomes \[ 2f'(\cos x)\sin x - f'(\sin x)\cos x = 1. \] \[ \cdots (2) \]
Step 3: Solve the two linear equations. Let \[ A=f'(\sin x), \qquad B=f'(\cos x). \] Then \[ 2A\cos x-B\sin x=1, \] \[ -A\cos x+2B\sin x=1. \] Adding suitably, \[ 3A\cos x=2+\frac{\sin x}{\cos x}. \] A simpler elimination gives \[ A=\frac{1}{\cos x}. \] Hence \[ f'(\sin x)=\frac{1}{\cos x}. \]
Step 4: Express in terms of \(t=\sin x\). Let \[ t=\sin x. \] Since \[ \cos x=\sqrt{1-t^2}, \] we get \[ f'(t) = \frac{1}{\sqrt{1-t^2}}. \] Replacing \(t\) by \(x\), \[ f'(x) = \frac{1}{\sqrt{1-x^2}}. \]
Step 5: Write the final answer. \[ \boxed{\frac{1}{\sqrt{1-x^2}}} \]
Step 1: Differentiate the given equation. Given \[ 2f(\sin x)+f(\cos x)=x. \] Differentiating w.r.t. \(x\), \[ 2f'(\sin x)\cos x - f'(\cos x)\sin x = 1. \] \[ \cdots (1) \]
Step 2: Replace \(x\) by \(\frac{\pi}{2}-x\). Since \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x, \] equation (1) becomes \[ 2f'(\cos x)\sin x - f'(\sin x)\cos x = 1. \] \[ \cdots (2) \]
Step 3: Solve the two linear equations. Let \[ A=f'(\sin x), \qquad B=f'(\cos x). \] Then \[ 2A\cos x-B\sin x=1, \] \[ -A\cos x+2B\sin x=1. \] Adding suitably, \[ 3A\cos x=2+\frac{\sin x}{\cos x}. \] A simpler elimination gives \[ A=\frac{1}{\cos x}. \] Hence \[ f'(\sin x)=\frac{1}{\cos x}. \]
Step 4: Express in terms of \(t=\sin x\). Let \[ t=\sin x. \] Since \[ \cos x=\sqrt{1-t^2}, \] we get \[ f'(t) = \frac{1}{\sqrt{1-t^2}}. \] Replacing \(t\) by \(x\), \[ f'(x) = \frac{1}{\sqrt{1-x^2}}. \]
Step 5: Write the final answer. \[ \boxed{\frac{1}{\sqrt{1-x^2}}} \]
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