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The instructions: Use the definition of derivative to find $f'(x)$ if $f(x)=\tan^2(x)$.

I've been working on this problem, trying every way I can think of. At first I tried this method: $$\lim_{h\to 0} {\tan^2(x+h)-\tan^2(x)\over h}$$ $$\lim_{h\to 0} {\tan(x+h)-\tan(x)\over h}\cdot\lim_{h\to 0} {\tan(x+h)-\tan(x)\over h}$$ And then I went on from there, but I was never able to get rid of the $h$.

So then I tried this: $$\lim_{x\to y} {\tan(x)-\tan(y)\over x-y}\cdot\lim_{x\to y} {\tan(x)-\tan(y)\over x-y}$$ $$\lim_{x\to y} {\tan(x-y)[1+\tan(x)\tan(y)]\over x-y}\cdot\lim_{x\to y} {\tan(x-y)[1+\tan(x)\tan(y)]\over x-y}$$ $$\lim_{x\to y} {\sin(x-y)\over (x-y)}\cdot{1\over \cos(x-y)}\cdot[1+\tan(x)\tan(y)]\cdot\lim_{x\to y} {\sin(x-y)\over (x-y)}\cdot{1\over \cos(x-y)}\cdot[1+\tan(x)\tan(y)]$$ I then put the ${\frac{\sin(x-y)}{x-y}}=1$, and I traded all of the $y$ values for $x$, which gave me $\frac{1}{\cos(\theta)}=1$ $$1+\tan^2(x)\cdot1+\tan^2(x)=1+2\tan^2(x)+\tan^4(x)$$ I know that the derivative of $\tan^2(x)=2\tan(x)\sec^2(x)$, so this is obviously wrong.

I found this link from a previous stack exchange point on finding the limit by definition of $\tan(x)$, so I tried using the answers given there. But, even with that, I haven't been able to get this correct.

What do I need to do?

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  • $\begingroup$ You''re using the identity $a^2-b^2=(a-b)^2$. But that identity is wrong. Think back to algebra. There is an important identity here; $a^2-b^2=$ what? $\endgroup$ Jul 25, 2015 at 19:06
  • $\begingroup$ And even the "identity" $\frac{a^2-b^2}h=\left(\frac{a-b}h\right)^2$... which is most bizarre. $\endgroup$
    – Did
    Jul 25, 2015 at 20:59
  • $\begingroup$ Yes, I see now that it's very wrong. $\endgroup$
    – matryoshka
    Jul 25, 2015 at 21:13

3 Answers 3

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Using first principle, the derivative of any function $f(x)$ is given as $$\frac{d(f(x))}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ Hence, derivative of $\tan^2 x$ is given as $$\frac{d(\tan^2 x)}{dx}=\lim_{h\to 0}\frac{\tan^2(x+h)-\tan^2(x)}{h}$$ $$=\lim_{h\to 0}\frac{(\tan(x+h)-\tan(x))(\tan(x+h)+\tan(x))}{h}$$$$=\lim_{h\to 0}\frac{\tan(x+h)-\tan(x)}{h}\times \lim_{h\to 0}(\tan(x+h)+\tan(x))$$ $$=\lim_{h\to 0}\frac{\frac{\sin(x+h)}{\cos(x+h)}-\frac{\sin(x)}{\cos(x)}}{h}\times \lim_{h\to 0}(\tan(x+h)+\tan(x))$$ $$=\lim_{h\to 0}\frac{\sin(x+h)\cos x-\cos(x+h)\sin x}{h\cos(x+h)\cos x}\times (\tan x+\tan x)$$ $$=2\tan x\lim_{h\to 0}\frac{\sin(x+h-x)}{h}\times \lim_{h\to 0}\frac{1}{\cos(x+h)\cos x}$$ $$=2\tan x\lim_{h\to 0}\frac{\sin h}{h}\times \frac{1}{\cos^2 x}$$ $$=2\tan x \times 1\times \sec^2 x$$ $$=\color{blue}{2\tan x \sec^2x}$$

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  • $\begingroup$ On step 6, $sin(x+h-x)$ was found by using the difference trig formula for sine, right? $\endgroup$
    – matryoshka
    Jul 25, 2015 at 19:50
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    $\begingroup$ yes, you are right $\endgroup$ Jul 25, 2015 at 19:53
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Your first way to try to solve was good, you were just a little off

$$\lim_{h\to 0}\frac{\tan^2(x+h)-\tan^2x}{h}=\lim_{h\to 0}\frac{(\tan(x+h)-\tan x)(\tan(x+h)+\tan x)}{h}$$

$$=\lim_{h\to0}\frac{\tan(x+h)-\tan x}{h}\cdot \lim_{h\to 0}({\tan(x+h)+\tan x})$$

$$=\frac{d}{dx}(\tan x)\cdot 2\tan x$$

$$=2\sec^2 x\tan x$$

This is of course assuming you're allowed to use the derivative of tangent.

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  • $\begingroup$ since $\tan(x)=\frac{\sin(x)}{\cos(x)}$ you have to lost some words about the points where $\cos(x)=0$ or not? $\endgroup$ Jul 25, 2015 at 19:16
  • $\begingroup$ Unfortunately, I'm not allowed to use the derivative of tangent. $\endgroup$
    – matryoshka
    Jul 25, 2015 at 19:19
  • $\begingroup$ then you can write $\tan(x)=\frac{\sin(x)}{\cos(x)}$ or you derive $\tan(x)'$ and then for the square $\endgroup$ Jul 25, 2015 at 19:24
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i think your derivative (with brackets) is ok since $\tan(x)$ has a derivative thus $\tan(x)\tan(x)=\tan(x)^2$ has also a derivative and is given by $2\cdot(1+\tan(x)^2)(\tan(x))$ (in the range of definition) a comment before the downvoting is starting: $$\tan(x)'=\frac{1}{\cos(x)^2}=\frac{\sin(x)^2+\cos(x)^2}{\cos(x)^2}=1+\tan(x)^2$$

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  • $\begingroup$ "i think your derivative (with brackets) is ok" ?? Are you doing this on purpose, Dr.? $\endgroup$
    – Did
    Jul 25, 2015 at 21:01
  • $\begingroup$ sorry i have misread it $\endgroup$ Jul 26, 2015 at 6:59

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