Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Mark the Correct Alternative in of the Following: If Y = Sin ( X + 9 ) Cos X Then D Y D X at X = 0 is - Mathematics

MCQ

Mark the correct alternative in  of the following:
If \[y = \frac{\sin\left( x + 9 \right)}{\cos x}\] then \[\frac{dy}{dx}\] at x = 0 is 

Options

  •  cos 9     

  • sin 9   

  •  0     

  • 1

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Solution

\[y = \frac{\sin\left( x + 9 \right)}{\cos x}\] 

Differentiating both sides with respect to x, we get 

\[\frac{dy}{dx} = \frac{\cos x \times \frac{d}{dx}\sin\left( x + 9 \right) - \sin\left( x + 9 \right) \times \frac{d}{dx}\cos x}{\cos^2 x} \left( \text{ Quotient rule } \right)\]
\[ = \frac{\cos x \times \cos\left( x + 9 \right) - \sin\left( x + 9 \right) \times \left( - \sin x \right)}{\cos^2 x}\]
\[ = \frac{\cos\left( x + 9 \right)\cos x + \sin\left( x + 9 \right)\sin x}{\cos^2 x}\]
\[ = \frac{\cos\left( x + 9 - x \right)}{\cos^2 x}\]
\[ = \frac{\cos9}{\cos^2 x}\]
Putting x = 0, we get 

\[\left( \frac{dy}{dx} \right)_{x = 0} = \frac{\cos9}{\cos^2 0} = \cos9\] 

Thus, \[\frac{dy}{dx}\]  at x = 0 is cos 9.

Hence, the correct answer is option (a).

 

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 10 | Page 48
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