Find the derivative of y= sinx/2+cotx
Find the derivative of y=sinx2+cotxy=\sin{\frac{x}{2}}+\cot{x}y=sin2x+cotx
Solution:
y′=12cosx2−1sin2xy'=\frac12\cos{\frac {x}{2}}-\frac{1}{\sin^2{x}}y′=21cos2x−sin2x1
Answer:
y′=cosx2⋅(x2)′−1sin2x=12cosx2−1sin2xy'=\cos{\frac {x}{2}}\cdot(\frac x2)'-\frac{1}{\sin^2{x}}=\frac12\cos{\frac {x}{2}}-\frac{1}{\sin^2{x}}y′=cos2x⋅(2x)′−sin2x1=21cos2x−sin2x1 .
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