Differentiate the following with regard to x : y=(cosec x +2tanx)^3
y′=((cosecx+2tanx)3)′==3(cosecx+2tanx)2⋅(cosecx+2tanx)′=3(cosecx+2tanx)2⋅(1sinx+2tanx)′==3(cosecx+2tanx)2⋅(−1sin2xcosx+2cos2x)=3(cosecx+2tanx)2⋅(−cotxsinx+2cos2x)==3(cosecx+2tanx)2⋅(2sec2x−cotxcosecx)y' = {\left( {{{\left( {\cos ecx + 2\tan x} \right)}^3}} \right)^\prime } = \\=3{\left( {\cos ecx + 2\tan x} \right)^2} \cdot {\left( {\cos ecx + 2\tan x} \right)^\prime } = 3{\left( {\cos ecx + 2\tan x} \right)^2} \cdot {\left( {\frac{1}{{\sin x}} + 2\tan x} \right)^\prime } =\\ =3{\left( {\cos ecx + 2\tan x} \right)^2} \cdot \left( { - \frac{1}{{{{\sin }^2}x}}\cos x + \frac{2}{{{{\cos }^2}x}}} \right) = 3{\left( {\cos ecx + 2\tan x} \right)^2} \cdot \left( { - \frac{{cotx}}{{\sin x}} + \frac{2}{{{{\cos }^2}x}}} \right) =\\= 3{\left( {\cos ecx + 2\tan x} \right)^2} \cdot \left( {2{{\sec }^2}x - cotx\cos ecx} \right)y′=((cosecx+2tanx)3)′==3(cosecx+2tanx)2⋅(cosecx+2tanx)′=3(cosecx+2tanx)2⋅(sinx1+2tanx)′==3(cosecx+2tanx)2⋅(−sin2x1cosx+cos2x2)=3(cosecx+2tanx)2⋅(−sinxcotx+cos2x2)==3(cosecx+2tanx)2⋅(2sec2x−cotxcosecx)
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