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

1 Answer
Dec 14, 2017

# 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) #