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32 Answers
32 Answers
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2sec² (2x)Modal content
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d/dx(tanx)=sec²x d/dx(tan2x)=2sec²2xModal content
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Sec2xModal content
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To differentiate(taking with respect to x) this we can use chain rule:- d/dx(Tan2x)=2Sec²(2x)Modal content
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2sec^2(2x)Modal content
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d/dx[tan(2x)]=sec²2x.d/dx(2x) Since d/dx(2x)=2 d/dx[tan(2x)]=2sec²(2x)Modal content
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2sec^2(2x)Modal content
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=Tan2x =2sec^2(2x)Modal content
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2 sec2xModal content
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d/dx(tan2x)=sec²2x•2=2sec²2xModal content
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2 tanx sec^2 xModal content
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2sec²2xModal content
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tan2x=sin2x/cos2x sin2x.tan2x=cos2x.tanxsquare2x sex2x_cos2xModal content
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please chech the attached link for detailed solution.
https://docs.google.com/document/d/1S1fG4mMzFzsGTg5uTCBTifQECI4cNNxUTqJb0r42yEw/edit?usp=sharingModal content
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2x(degree)=[2*(22/7)*x]/180=(22x/630)
tan2x(degree)=tan(22x/630)
so the answer will be [(22/630][(sec2x)^2]
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we substtitute 2x=u, we get y=tanu,u=2x
dy/du=sec^2(u),du/dx=2
dy/dx=dy/du * du/dx (chain rule)
=sec^2(u) * 2Modal content
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d(tan 2x)/d(2x)* d(2x)/dx= 2sec^2 2xModal content
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if we substtitute 2x=u, we get y=tanu,u=2x
dy/du=sec^2(u),du/dx=2
dy/dx=dy/du * du/dx (chain rule)
=sec^2(u) * 2
=2sec^2(2x) (sub u=2x back in)Modal content
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Tan2x= (sec)2x*d2x/f2f =(sec)2*2Modal content
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