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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