KCET · Maths · Differentiation
The derivative of \(\sin \left(x^{3}\right)\) w.r.t. \(\cos \left(x^{3}\right)\) is
- A \(-\tan \left(\mathrm{x}^{3}\right)\)
- B \(\tan \left(\mathrm{x}^{3}\right)\)
- C \(-\cot \left(\mathrm{x}^{3}\right)\)
- D \(\cot \left(\mathrm{x}^{3}\right)\)
Answer & Solution
Correct Answer
(C) \(-\cot \left(\mathrm{x}^{3}\right)\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{u}=\sin \mathrm{x}^{3}\) and \(\mathrm{v}=\cos \mathrm{x}^{3}\).
On differentiating w.r.t. \(x\), we get \(\frac{\mathrm{du}}{\mathrm{dx}}=\cos \mathrm{x}^{3} \cdot 3 \mathrm{x}^{2} \quad\) and \(\quad \frac{\mathrm{dv}}{\mathrm{dx}}=-\sin \mathrm{x}^{3} \cdot 3 \mathrm{x}^{2}\) \(\therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{3 x^{2} \cos x^{3}}{-3 x^{2} \sin x^{3}}=-\cot x^{3}\)
On differentiating w.r.t. \(x\), we get \(\frac{\mathrm{du}}{\mathrm{dx}}=\cos \mathrm{x}^{3} \cdot 3 \mathrm{x}^{2} \quad\) and \(\quad \frac{\mathrm{dv}}{\mathrm{dx}}=-\sin \mathrm{x}^{3} \cdot 3 \mathrm{x}^{2}\) \(\therefore \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{3 x^{2} \cos x^{3}}{-3 x^{2} \sin x^{3}}=-\cot x^{3}\)
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