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Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?
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Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?
Deriving the Derivative of Y
Given function: Y = tan^(-1)((acosx - bsinx)/(bcosx + asinx))
Step 1: Simplify the expression
Using trigonometric identities, we can simplify the expression:
Y = tan^(-1)((acosx - bsinx)/(bcosx + asinx))
Y = tan^(-1)((acosx - bsinx)/(bcosx + asinx))
Y = tan^(-1)((a - b tanx)/(b + a tanx))
Step 2: Differentiate using Quotient Rule
To find dy/dx, we need to differentiate Y with respect to x using the Quotient Rule:
dy/dx = [(b + a tanx)(1) - (a - b tanx)(sec^2x)] / (b + a tanx)^2
dy/dx = [(b + a tanx) - (a - b tanx sec^2x)] / (b + a tanx)^2
dy/dx = [b + a tanx - a sec^2x + b tanx sec^2x] / (b + a tanx)^2
Step 3: Simplify the derivative
dy/dx = (2ab sec^2x) / (b + a tanx)^2
dy/dx = 2ab sec^2x / (b + a tanx)^2
Final Result: dy/dx = 2ab sec^2x / (b + a tanx)^2
This is the derivative of Y with respect to x.
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Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?
Questions wrong h... (bcosx +sinx) hoga denominator mai...
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Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?
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Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?.
Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Y = tan invrse of ( acosx - bsinx / bcosx asinx ) , dy/dx ?.
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