140k views
0 votes
Derivative of tan(2x+3) using first principle

User Jim Webber
by
9.0k points

1 Answer

4 votes

f(x)=\tan(2x+3)

The derivative is given by the limit


f'(x)=\displaystyle\lim_(h\to0)\frac{f(x+h)-f(x)}h

You have


\displaystyle\lim_(h\to0)\frac{\tan(2(x+h)+3)-\tan(2x+3)}h

\displaystyle\lim_(h\to0)\frac{\tan((2x+3)+2h)-\tan(2x+3)}h

Use the angle sum identity for tangent. I don't remember it off the top of my head, but I do remember the ones for (co)sine.


\tan(a+b)=(\sin(a+b))/(\cos(a+b))=(\sin a\cos b+\cos a\sin b)/(\cos a\cos b-\sin a\sin b)=(\tan a+\tan b)/(1-\tan a\tan b)

By this identity, you have


\tan((2x+3)+2h)=(\tan(2x+3)+\tan2h)/(1-\tan(2x+3)\tan2h)

So in the limit you get


\displaystyle\lim_(h\to0)\frac{(\tan(2x+3)+\tan2h)/(1-\tan(2x+3)\tan2h)-\tan(2x+3)}h

\displaystyle\lim_(h\to0)(\tan(2x+3)+\tan2h-\tan(2x+3)(1-\tan(2x+3)\tan2h))/(h(1-\tan(2x+3)\tan2h))

\displaystyle\lim_(h\to0)(\tan2h+\tan^2(2x+3)\tan2h)/(h(1-\tan(2x+3)\tan2h))

\displaystyle\lim_(h\to0)\frac{\tan2h}h*\lim_(h\to0)(1+\tan^2(2x+3))/(1-\tan(2x+3)\tan2h)

\displaystyle\frac12\lim_(h\to0)\frac1{\cos2h}*\lim_(h\to0)(\sin2h)/(2h)*\lim_(h\to0)(\sec^2(2x+3))/(1-\tan(2x+3)\tan2h)

The first two limits are both 1, and the single term in the last limit approaches 0 as
h\to0, so you're left with


f'(x)=\frac12\sec^2(2x+3)

which agrees with the result you get from applying the chain rule.
User Arif Amirani
by
7.1k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.