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If y = 1 + sinx - cosx ÷ 1 + sinx + cosx , then find dy/dx

Jashandeep , 9 Years ago
Grade 12
anser 1 Answers
Vikas TU
=d/dx[sin(x)−cos(x)+1]⋅(sin(x)+cos(x)+1)−(sin(x)−cos(x)+1)⋅d/dx[sin(x)+cos(x)+1]/(sin(x)+cos(x)+1)2
=(d/dx[sin(x)]−d/dx[cos(x)]+d/dx[1])(sin(x)+cos(x)+1)−(d/dx[sin(x)]+d/dx[cos(x)]+d/dx[1])/(sin(x)−cos(x)+1)(sin(x)+cos(x)+1)2
=(cos(x)−(−sin(x))+0)(sin(x)+cos(x)+1)−(cos(x)+(−sin(x))+0)(sin(x)−cos(x)+1)/(sin(x)+cos(x)+1)2
=(sin(x)+cos(x))(sin(x)+cos(x)+1)−(cos(x)−sin(x))(sin(x)−cos(x)+1)/(sin(x)+cos(x)+1)2
=sin(x)+cos(x)sin(x)+cos(x)+1−(cos(x)−sin(x))(sin(x)−cos(x)+1)/(sin(x)+cos(x)+1)2
 
=2(sin2(x)+sin(x)+cos2(x))/(sin(x)+cos(x)+1)2.
Last Activity: 8 Years ago
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