# Thread: Using first principles, find the derivative of y = tanx.

1. ## Using first principles, find the derivative of y = tanx.

I am rather terrible at using trigonometric proofs, which is needed in doing this question.

Question is title of this thread.

My attempt:
$\displaystyle \frac{tan(x+h) - tanx}{h}$
$\displaystyle = \frac{\frac{sin(x+h)}{cos(x+h)} - \frac{sinx}{cosx}}{h}$
$\displaystyle = \frac{\frac {sinxcosh + cosxsina}{cosxcosh - sinxsinh} - \frac{sinx}{cosx}}{h}$

I don't know what to do from here. I need to show all my steps to get the answer. I know the answer is $\displaystyle sec^2 x$, but don't know how to show it using first principles.

If you could show me what to do next, I would be very grateful.

2. Originally Posted by Kakariki
I am rather terrible at using trigonometric proofs, which is needed in doing this question.

Question is title of this thread.

My attempt:
$\displaystyle \frac{tan(x+h) - tanx}{h}$
$\displaystyle = \frac{\frac{sin(x+h)}{cos(x+h)} - \frac{sinx}{cosx}}{h}$
$\displaystyle = \frac{\frac {sinxcosh + cosxsina}{cosxcosh - sinxsinh} - \frac{sinx}{cosx}}{h}$

I don't know what to do from here. I need to show all my steps to get the answer. I know the answer is $\displaystyle sec^2 x$, but don't know how to show it using first principles.

If you could show me what to do next, I would be very grateful.
Use this identity: $\displaystyle \tan(x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}$

So $\displaystyle \tan(x+h)-\tan(x)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}-\tan(x)=$ $\displaystyle \frac{\tan(x)+\tan(h)-\tan(x)+\tan^2(x)\tan(h)}{1-\tan(x)\tan(h)}=\frac{\tan(h)(1+\tan^2(x))}{1-\tan(x)\tan(h)}$

Thus, $\displaystyle \lim_{h\to0}\frac{\tan(x+h)-\tan(x)}{h}=\lim_{h\to0}\frac{\tan(h)(1+\tan^2(x)) }{h(1-\tan(x)\tan(h))}$

Spoiler:
$\displaystyle \lim_{h\to0}\frac{\tan(h)(1+\tan^2(x))}{h(1-\tan(x)\tan(h))}=\lim_{h\to0}\left(\frac{\sin(h)}{ h}\cdot\frac{\frac{1}{\cos(h)}(1+\tan^2(x))}{1-\tan(x)\tan(h)}\right)=$ $\displaystyle \left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)\cdot\left(\lim_{h\to0}\ frac{\frac{1}{\cos(h)}(1+\tan^2(x))}{1-\tan(x)\tan(h)}\right)$ $\displaystyle =(1)\cdot\left(\frac{1\cdot(1+\tan^2(x))}{1-1\cdot 0}\right)=1+\tan^2(x)$

By a trig identity, $\displaystyle 1+\tan^2(x)=\boxed{\sec^2(x)}$

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