ysecx+2tanx+x2y=0

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