Question

What is the derivative of tanx?

Answer 1

see below

Explanation:

one approach is to use the quotient rule

`d/(dx)(u/v)=(vu'-uv')/v^2`

`d/(dx)(tanx)=d/(dx)(sinx/cosx)`

`=(cosx(d/(dx)(sinx))-sinx(d/(dx)(cosx)))/(cos^2x)`

`=(cosxcosx-sinx(-sinx))/cos^2x`

`=(cos^2x- -sin^2x)/cos^2x`

`=(cos^2x+sin^2x)/cos^2x`

`=1/cos^2x=sec^2x`

Answer 2

`dy/dx = sec^2x`

Explanation:

If:

`y = tanx`

Then applying the quotient rule we have:

`dy/dx = d/dx(sinx/cosx)`
`\ \ \ \ \ = ( (cosx)(d/dx sinx) - (sinx)(d/dx cosx) )/ (cosx)^2`
`\ \ \ \ \ = ( (cosx)(cosx) - (sinx)(-sinx) )/ (cos^2x)`
`\ \ \ \ \ = ( cos^2x + sin^2x )/ (cos^2x)`
`\ \ \ \ \ = ( 1 )/ (cos^2x)`
`\ \ \ \ \ = sec^2x`

Answer 3

see below

Explanation:

we could do it from first principles

`(dy)/(dx)=Lim_(hrarr0)(f(x+h)-f(x))/h`

for `y=tanx`

`(dy)/(dx)=Lim_( hrarr0)((tan(x+h)-tanx))/h`

`(dy)/(dx)=Lim_( hrarr0)((tan(x+h)-tanx))/h`

`(dy)/(dx)=Lim_( hrarr0)(((tanx+tanh)/(1-tanxtanh)-tanx))/h`

`(dy)/(dx)=Lim_( hrarr0)((cancel(tanx)+tanh-cancel(tanx)+tan^2xtanh))/(h(1-tanxtanh))`

`(dy)/(dx)=Lim_( hrarr0)(tanh+tan^2xtanh)/(h(1-tanxtanh))`

`(dy)/(dx)=Lim_( hrarr0)(tanh(1+tan^2x)/(h(1-tanxtanh)))`

we need then to evaluate the limit of

`Lim_(hrarr0)(tanh/(h(1-tanxtanh)))`

`now Lim_(hrarr0)(1-tanxtanh)=1-0=1`

so

`Lim_(hrarr0)(tanh/(h(1-tanxtanh)))=Lim_(hrarr0)(tanh/h)`

`=Lim_(hrarr0)(sinh/h1/cosh)`

`=Lim_(hrarr0)(sinh/h)xxLim_(hrarr0)1/cosh)`

`=1xx1=1`

`(dy)/(dx)=1+tan^2x=sec^2x`