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  • Please help! - Limit , continuity and differentiability - JEE Main-13

Let y= \sin ^{2}x + 2\cos ^{3}2x  then dy/dx equals 

  • Option 1)

    \sin 2x + 12 \sin 2x \cos ^{2}2x

  • Option 2)

    \cos^{2x }+ 6 \cos ^{2}2x

  • Option 3)

    \sin 2x - 12 \sin 2x \cos ^{2}2x

  • Option 4)

    \cos^{2x }+ 6 \sin ^{2}2x

 

Answers (1)

best_answer

As we have learned

Rule for differentiation -

The derivative of the sum or difference of two functions is the sum or difference of their derivatives.

\frac{d}{dx}\:{f(x)\pm g(x)}=\frac{d}{dx}\:{f(x)}\pm \frac{d}{dx}\:{g(x)}

-

 

 y= \sin^{2}x+2\cos ^{3}2x \Rightarrow \frac{dy}{dx}= \frac{d}{dx}( \sin^{2}x+2\cos ^{3}2x)

=\frac{d}{dx} \sin^{2}x+\frac{d}{dx}2\cos ^{3}2x = \frac{d}{dx} (\sin x)^{2} +2 \frac{d}{dx}(\cos 2x )^{3}

=\frac{d(\sin x)}{dx} \frac{d(\sin^{2}x)}{d(\sin x)} +2\frac{d(\cos 2x )^{3}}{d(\cos 2x)}\cdot \frac{d(2x)}{dx}

= 2\sin x\cos x+ 2*3(\cos x)^{2}* (-sin2x)*2

= \sin 2x -12 \sin 2x\cos^{2} x

 

 

 

 

 


Option 1)

\sin 2x + 12 \sin 2x \cos ^{2}2x

Option 2)

\cos^{2x }+ 6 \cos ^{2}2x

Option 3)

\sin 2x - 12 \sin 2x \cos ^{2}2x

Option 4)

\cos^{2x }+ 6 \sin ^{2}2x

Posted by

Himanshu

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