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Calculus1
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Find dy/dx such that x=pi/3 a) csc(pi-x) b) 2^tanx
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d(cos(pi-x))/dx = -sin(pi-x) * d(-x)/dx =-sin(pi-x)*(-1) =sin(pi-x) at x= pi/3 sin(pi-pi/3)=(3^1/2)/2
\[y=2^(tanx)\]
take log both side log y=tanx log2 dfferentiate (1/y )*dy/dx=sec^2x * ln2 dy/dx=y *sec^2x * ln2 but y= 2^tanx dy/dx=(2^tanx)*sec^2x * ln2 at x=pi/3 dy/dx = \[2^\sqrt{3}*2^2 * \ln2\]
\[2.772*2^\sqrt{3}\]
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