Math, asked by divyanshi1813, 10 months ago

tan^{-1} (3x - x^{2}/ 1 - 3x^{2}) dy/dx ज्ञात कीजिए

Answers

Answered by amitnrw
0

dy/dx =  (9x²  -2x + 3)/(9x⁴ - 7x²   + 3x + 1) यदि y = Tan⁻¹((3x - x²)/(1 - 3x²))

Step-by-step explanation:

dy/dx ज्ञात कीजिए

y = Tan⁻¹((3x - x²)/(1 - 3x²))

=> Tany  = (3x - x²)/ (1 - 3x²)

=> Sec²y (dy/dx)  =     ((3x - x²)(-1)/(1 - 3x²)²)(-6x)  + (3 - 2x)/(1 - 3x²)

=> (1 + Tan²y)(dy/dx)  =  (18x² - 6x³)/(1 - 3x²)²  +  (3 - 2x)/(1 - 3x²)

=> (1 + Tan²y)(dy/dx)  =  (18x² - 6x³ )/(1 - 3x²)²  +  (3 - 2x)(1 - 3x²)/(1 - 3x²)²

=> (1 + Tan²y)(dy/dx)  =  (18x² - 6x³ +   6x³  + 3 - 2x  - 9x²)/(1 - 3x²)²

=> (1 + Tan²y)(dy/dx)  =  (9x²  -2x + 3)/(1 - 3x²)²

1 + Tan²y  = 1   +  ((3x - x²)/ (1 - 3x²))²

=> 1 + Tan²y  =  (1  + 9x⁴ - 6x²   + 3x  - x²)/(1 - 3x²)²

=> 1 + Tan²y  =  (9x⁴ - 7x²   + 3x + 1)/(1 - 3x²)²

(9x⁴ - 7x²   + 3x + 1)/(1 - 3x²)²(dy/dx)  =  (9x²  -2x + 3)/(1 - 3x²)²

=> dy/dx =  (9x²  -2x + 3)/(9x⁴ - 7x²   + 3x + 1)

और अधिक जानें :

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