find the first derivative for y=3/(cosx) + 2/(tanx)

Question

find the first derivative  for  y=3/(cosx) + 2/(tanx)

shivaniits · Answer

$$y=3\sec x+2\cot x$$
$$dy/dx=(3\tan x \times \sec x)-2(\csc x)^{2}$$

anonymous · Answer

What is the rule to figure this out. Because I thought that the derivative of cosx is -sinx and the derivative of tanx is sec^2x

Do you do the quotient rule for both numbers first and then do the sum rules?
: 0(cosx)-3(-sinx) = 3sinx
0(tanx)-2(sec^2x) = -2sec^2x
Am I doing it right so far?

anonymous · Answer

Oh nevermind I just saw that the derivative of 1/cosx is tanxsecx
and the derivative of 1/tanx is -csc^2x

Thank you so much for your help

shivaniits · Answer

how to figure out the derivative of cot x
$$f(x)=\cot x=\cos x/\sin x$$
$$f'(x)=[sinx (-sinx)-cosx(cosx)]/(sinx)^2$$ 
$$f'(x)=[-\sin^2x-\cos^2x]/(sinx)^2$$
$$f'(x)=-(\sin^2x +\cos^2x)/\sin^2x$$
$$f'(x)=-(1/\sin^2 x)$$ 
$$f'(x)=-\csc^2x  $$

another solution 
$$f(x)=\cot x=1/tanx  $$
$$f'(x)=[tanx(0)-1(\sec^2x)]/\tan^2x  $$
$$f'(x)=-\sec^2x/\tan^2x  $$
$$f'(x)=-(1/\cos^2x)(\cos^2x/\sin^2x)  $$
$$f'(x)=-1/\sin^2x=-\csc^2x $$
how to figure out the derivative for sec x
$$secx=1/cosx  $$
$$(secx)'=[1'cosx-1.cosx']/cosx²  $$
$$(secx)'=[0.cosx-1.(-sinx)]/cosx² $$ 
$$(secx)'=[sinx]/cosx²  $$
$$(secx)'=[sinx]/cosx].1/cosx  $$
$$(secx)'= \tan x \times \sec x$$