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

Question

Emilia Aguilar · Accepted Answer

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

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) #

Adam Adkins · Answer

undefined To find the derivative of $ \tan(ax + b) $ from first principles, we start with the definition of the derivative:

$$ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h} $$

Let's apply this definition to $ \tan(ax + b) $:

$$ f(x) = \tan(ax + b) $$
$$ f'(x) = \lim_{{h \to 0}} \frac{{\tan(a(x + h) + b) - \tan(ax + b)}}{h} $$

Next, we use the trigonometric identity for the difference of tangents:

$$ \tan(x - y) = \frac{{\tan x - \tan y}}{{1 + \tan x \tan y}} $$

Applying this identity to our expression:

$$ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{\tan(a(x + h) + b) - \tan(ax + b)}}{{1 + \tan(a(x + h) + b) \tan(ax + b)}}}}{h} $$

Now, we simplify the expression:

$$ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{\frac{{\sin(a(x + h) + b)}}{{\cos(a(x + h) + b)}} - \frac{{\sin(ax + b)}}{{\cos(ax + b)}}}}{{1 + \frac{{\sin(a(x + h) + b)}}{{\cos(a(x + h) + b)}} \cdot \frac{{\sin(ax + b)}}{{\cos(ax + b)}}}}}}{h} $$

$$ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{\sin(a(x + h) + b) \cdot \cos(ax + b) - \sin(ax + b) \cdot \cos(a(x + h) + b)}}{{\cos(a(x + h) + b) \cdot \cos(ax + b)}}}}{{h \cdot (\cos(a(x + h) + b) \cdot \cos(ax + b) + \sin(a(x + h) + b) \cdot \sin(ax + b))}} $$

After simplifying and factoring, we obtain:

$$ f'(x) = \lim_{{h \to 0}} \frac{{a(\cos(ax + b) \cdot \cos(a(x + h) + b) - \sin(ax + b) \cdot \sin(a(x + h) + b))}}{{h \cdot (\cos(a(x + h) + b) \cdot \cos(ax + b) + \sin(a(x + h) + b) \cdot \sin(ax + b))}} $$

By using trigonometric identities and limits, we further simplify the expression:

$$ f'(x) = a \cdot \frac{{\cos^2(ax + b) - \sin^2(ax + b)}}{{(\cos(ax + b))^2 + (\sin(ax + b))^2}} $$

$$ f'(x) = a \cdot \frac{{\cos^2(ax + b) - (1 - \cos^2(ax + b))}}{{\cos^2(ax + b) + (1 - \cos^2(ax + b))}} $$

$$ f'(x) = a \cdot \frac{{2\cos^2(ax + b) - 1}}{{2\cos^2(ax + b)}} $$

$$ f'(x) = a \cdot \frac{{\cos(2(ax + b))}}{{\cos^2(ax + b)}} $$

Therefore, the derivative of $ \tan(ax + b) $ from first principles is $ a \cdot \frac{{\cos(2(ax + b))}}{{\cos^2(ax + b)}} $.