ysecx+2tanx+x2y=0
Asked by Pranshu Chadha
| 17th Nov, 2012,
01:37: PM
Expert Answer:
Answer : Given : ysecx +2tanx +x2y =0
To find dy /dx
=> ysecx +2tanx +x2y =0
differentiate wrt x
=> y (secxtanx) + secx (dy/dx) + 2 sec2x + 2xy +x2 (dy/dx)=0 {using product rule and d{x2} / dx = 2x and d{secx}/ dx = secx tanx and d{tanx}/dx = sec2x}
=> y secx tanx + 2 sec2x + 2xy=-(secx +x2) (dy/dx)
=> dy/dx = -[ y secx tanx + 2 sec2x + 2xy] / [secx +x2] Answer
Answered by
| 17th Nov, 2012,
04:16: PM
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