Question

Find the derivative of `tan(ax+b)` from first principles?

Answer 1

`d/dx tan(ax+b) = asec^2(ax+b)`

Explanation:

Using the derivative definition, if:

`f(x) = tan(ax+b)`

Then, the derivative `f'(x)` is given by:

`f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) )/ h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( tan(a(x+h)+b)-tan(ax+b) ) /h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( tan((ax+b)+ah)-tan(ax+b) ) /h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (tan (ax+b)+tan ah)/(1-tan (ax+b) tan ah )-tan(ax+b) ) /h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (tan (ax+b)+tan ah - tanx(1-tan (ax+b) tan ah )) / ( 1-tan (ax+b) tan ah ) ) /h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( tan (ax+b)+tan ah - tan(ax+b)+tan^2 (ax+b) tan ah ) / (h( 1-tan (ax+b) tan ah ) )`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( tan ah (1+ tan^2(ax+b)) ) / (h( 1-tan (ax+b) tan ah ) )`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 1+ tan^2(ax+b)) / ( 1-tan (ax+b) tan ah ) * (tan ah)/h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 1+ tan^2(ax+b)) / ( 1-tan (ax+b) tan ah ) * lim_(h rarr 0) (tan ah)/h`

Consider the first limit:

`L_1 = lim_(h rarr 0) ( 1+ tan^2(ax+b)) / ( 1-tan (ax+b) tan ah )`
`\ \ \ \ = ( 1+ tan^2(ax+b)) / ( 1-tan (ax+b) 0 )`
`\ \ \ \ = 1+ tan^2(ax+b)`
`\ \ \ \ = sec^2(ax+b)`

And, now the second limit:

`L_2 = lim_(h rarr 0) (tan ah)/h`
`\ \ \ \ = lim_(h rarr 0) (sin ah)/(cos ah) * 1/h`
`\ \ \ \ = lim_(h rarr 0) (sin ah)/h * 1/(cos ah)`
`\ \ \ \ = lim_(h rarr 0) (asin ah)/(ah) * 1/(cos ah)`
`\ \ \ \ = lim_(h rarr 0) (asin ah)/(ah) * lim_(h rarr 0) 1/(cos ah)`
`\ \ \ \ = a \ lim_(theta rarr 0) (sin theta)/(theta) * lim_(h rarr 0) 1/(cos ah)`

And for this limit we have:

`lim_(theta rarr 0) (sin theta)/(theta) =1` and `lim_(h rarr 0) 1/(cos ah) = 1`

Leading to:

`L_2 = a`

Combining these results we have:

`f'(x) = sec^2(ax+b) * a`
`\ \ \ \ \ \ \ \ \ = asec^2(ax+b)`