Question

Derivative of tan(2x+3) using first principle

Answer 1

`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.