# The derivative of $f(x) = \tan x\sec x$ using first principles

I have to find out the derivative of given function by using 'first principle method', but the question has eaten half of my brain, the problem is first principle method. The function is $f(x)= \tan x\sec x$.

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Is this a homework question? The problem likely wants you to find the derivative of the function using the limit definition of the derivative. – JavaMan Jan 4 '13 at 21:48
It is from my test sheet. – Syed Sahl Jan 4 '13 at 21:52
Well, you shouldn't be asking for help if it is on a test... – Zev Chonoles Jan 4 '13 at 21:59
@ZevChonoles Often a "test sheet" is a sheet of problem to review before a test. – Alex Becker Jan 4 '13 at 22:00
@Alex: Hmm, I never heard that term before. Sorry for making that assumption, Syed! – Zev Chonoles Jan 4 '13 at 22:05

Because they are more familiar, we express $f(x)$ in terms of sines and cosines. We get $$f(x)=\frac{\sin x}{\cos^2 x}=\frac{\sin x}{1-\sin^2 x}.$$ Thus $$f'(x)=\lim_{h\to 0} \frac{\frac{\sin(x+h)}{1-\sin^2(x+h)}-\frac{\sin x}{1-\sin^2 x}}{h}.$$ We will need to impose the condition $\sin x\ne \pm 1$. Bring the top to the common denominator $(1-\sin^2(x+h))(1-\sin^2 x)$. Some algebra shows that we want $$\lim_{h\to 0}\frac{1+\sin(x+h)\sin x}{(1-\sin^2(x+h))(1-\sin^2 x)} \frac{\sin(x+h)-\sin x}{h}.$$

Now copy the standard first principles argument that shows that $\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}=\cos x$. The limits of the remaining terms are easy to compute.

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But that's what I have. – André Nicolas Jan 4 '13 at 23:08
My shift key sticks. So I end up having 62 fairly often instead of ^2. – André Nicolas Jan 4 '13 at 23:14
I have a similar problem! (different key, but same sort of sticky-key problem!) – amWhy Jan 4 '13 at 23:15

Hint: I will map out the approach and see if you can follow it.

Definition: The derivative of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ at a value $x = a$, if it exists, is $$\displaystyle \lim_{h \rightarrow 0} = \large \frac{f(a + h) - f(a)}{h}$$

So, you have $f(x) = \tan(x) \sec(x)$

Let $u(x) = \tan(x)$ and $v(x) = sec(x)$?

Using first principles, you can derive the equation $f'(x)$ for $f(x) = u(x)v(x)$, because we have:

$$\displaystyle \lim_{h \rightarrow 0} = \large \frac{f(a + h) - f(a)}{h} = \frac{u(a + h)v(a + h) - u(a)v(a)}{h}$$

After expanding this and taking the limit, we see that it reduces to $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Do you see that this is the product rule? Do you follow this? Can you actually derive it?

Now, we need $u'(x)$ and $v'(x)$ for $\tan(x)$ and $sec(x)$ and here we will make use of first principles again for those derivatives.

Find the derivative of $\tan(x)$ using first principles.

Using the identity $\sin(A)\cos(B) - \cos(A)\sin(B) = \sin(A-B)$, we find

$\tan(x+h) - \tan(x)$

$= \large \frac{\sin(x+h)}{\cos(x+h)} - \frac{\sin(x)}{\cos(x)}$

$= \large \frac{\sin(x+h)\cos(x)-\cos(x+h)\sin(x)}{\cos(x+h)\cos(x)}$

$= \large \frac{\sin(x+h-x)}{\cos(x+h)\cos(x)}$

$= \large \frac{\sin(h)}{\cos(x+h)\cos(x)}$

$\Rightarrow \large \frac{\tan(x+h) - \tan(x)}{h} = \frac{\sin(h)}{h}[\frac{1}{\cos(x+h)\cos(x)}]$

$\Rightarrow \large \displaystyle \lim_{h \rightarrow 0} \frac{\tan(x+h) - \tan(x)}{h} = \frac{1}{\cos^{2}(x)} = \sec^{2}(x)$

Can you finish it off? Hint: Find the derivative of $\sec(x)$ using first principles and then use the product rule we derived above.

The final result should be: $$f'(x) = \sec(x) (\tan^{2}(x) + \sec^{2}(x))$$

Regards

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