First u just differentiate sin^2(ø^2+1) with respect to sin(ø^2+1) so we get 2sin(ø^2+1)..Then again differentiate sin(ø^2+1) with respect to (ø^2+1) so we get cos(ø^2+1). and now lastly we differentiate (ø^2+1) with respect to ø^2 so we get 1 ..... and lastly the result is 2sin(ø^2+1)×cos(ø^2+1)×1 = sin2(ø^2+1) = sin(2ø^2+2)..... Here , ø= theta ....
Differentiate sin2(θ2 + 1) with respect toθ2a)sin(2θ2 + 1)b)cos(2θ2 + 2)c)sin(2θ2 + 2)d)cos(2θ2 + 1)Correct answer is option 'C'. Can you explain this answer?
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Differentiate sin2(θ2 + 1) with respect toθ2a)sin(2θ2 + 1)b)cos(2θ2 + 2)c)sin(2θ2 + 2)d)cos(2θ2 + 1)Correct answer is option 'C'. Can you explain this answer? for JEE 2023 is part of JEE preparation. The Question and answers have been prepared
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Differentiate sin2(θ2 + 1) with respect toθ2a)sin(2θ2 + 1)b)cos(2θ2 + 2)c)sin(2θ2 + 2)d)cos(2θ2 + 1)Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Differentiate sin2(θ2 + 1) with respect toθ2a)sin(2θ2 + 1)b)cos(2θ2 + 2)c)sin(2θ2 + 2)d)cos(2θ2 + 1)Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Differentiate sin2(θ2 + 1) with respect toθ2a)sin(2θ2 + 1)b)cos(2θ2 + 2)c)sin(2θ2 + 2)d)cos(2θ2 + 1)Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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First u just differentiate sin^2(ø^2+1) with respect to sin(ø^2+1) so we get 2sin(ø^2+1)..Then again differentiate sin(ø^2+1) with respect to (ø^2+1) so we get cos(ø^2+1). and now lastly we differentiate (ø^2+1) with respect to ø^2 so we get 1 ..... and lastly the result is 2sin(ø^2+1)×cos(ø^2+1)×1 = sin2(ø^2+1) = sin(2ø^2+2)..... Here , ø= theta ....
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