# Using first principles find derivative of ln(sec(x))

The question is to use first principles only.

Thus I started with the same and got

$$y = \ln(\sec(x))$$ $$\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln(\sec(x+h)) - \ln(\sec(x))}{h}$$

after this I do not understand how do I eliminate the $$h$$ in the denominator. I tried to implement $$\ln(A) - \ln(B) = \ln\bigl(\frac{A}{B}\bigr)$$ which ultimately led to

$$\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\sec(x+h)}{\sec(x)}\bigr)}{h}$$

here I converted $$\sec()$$ to $$\cos()$$

$$\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\cos(x)}{\cos(x+h)}\bigr)}{h}$$

Still I cannot proceed further.

• You could try using Taylor series expansion of cos(x) and see what happens. Aug 10, 2021 at 3:55
• Try using the trigonometric identities : $\cos(x+h) = \cos x \cos h - \sin x \sin h$.
– Vue
Aug 10, 2021 at 3:57
• socratic.org/questions/… Well here's a link that will solve your problems Aug 10, 2021 at 3:57
• well that link is for solving derivative of sec(x) but here it is ln(sec(x)) @nishant singh and using taylor series i can expand it but then how will i eliminate h in the denominator Aug 10, 2021 at 4:01
• following Claude's advice $$lim_{h=0} \frac{-ln(\frac{cos(x+h)}{cos(x)})}{h}$$ then using identity $$lim_{h=0} \frac{-ln(\frac{cos(x)cos(h) - sin(x)sin(h)}{cos(x)})}{h}$$ this converts that fraction into $$lim_{h=0} \frac{-ln(cos(h) - \frac{sin(x)sin(h)}{cos(x)})}{h}$$ It is not $ln(1-y)$ Aug 10, 2021 at 4:23

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\sec(x+h)) -\ln(\sec(x))}{h}$$

Using $$\ln(A) - \ln(B) = \ln(\frac{A}{B})$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\sec(x+h)}{\sec(x)})}{h}$$

coverting $$\sec(x)$$ to $$\cos(x)$$ using $$\cos(x) = \frac{1}{\sec(x)}$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\cos(x)}{\cos(x+h)})}{h}$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x)}{\cos(x+h)}-1)}{h}$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}$$

multiplying and dividing by $$\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}$$

repositioning

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}$$

seperating limit

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}$$

As $$h$$ approaches 0 so does $$\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$$ as the numerator beging to approach $$0$$. ($$\cos(x) - \cos(x)$$)

let us assume $$t = \frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$$ and hence t approaches $$0$$ when $$h$$ approaches $$0$$

hence equation turns out to be

$$\frac{d}{dx}\ln\sec(x) = \lim_{t\to0} \frac{\ln(1+t)}{t} \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}$$

Using the standard limit $$\lim_{x\to0} \frac{ln(x+1)}{x} = 1$$

Therefore

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\cos(x)-\cos(x+h)}{h\cos(x+h)}$$

Applying $$\cos(A) - \cos(B) = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$$

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{-2\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{h\cos(x+h)}$$

Bringing the $$-2$$ down

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{\frac{-h}{2}\cos(x+h)}$$

rearranging

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)} \lim_{h\to0} \frac{\sin(\frac{-h}{2})}{\frac{-h}{2}}$$

Using standard limit $$\lim_{x\to0} \frac{sin(x)}{x} = 1$$

Therefore

$$\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)}$$

Putting $$h = 0$$

$$\frac{d}{dx}\ln\sec(x) = \frac{\sin(\frac{2x}{2})}{\cos(x)}$$

$$\frac{d}{dx}\ln\sec(x) = \frac{\sin(x)}{\cos(x)}$$

$$\frac{d}{dx}\ln\sec(x) = \tan(x)$$

• There is also a possibility of using chain rule in terms of the first principles, look here. Aug 10, 2021 at 4:56

$$\displaystyle \frac{d}{dx} {\ln(\sec x)}=\lim_{h \to 0} \frac{-ln(\frac{\cos(x)\cos(h) - \sin(x)\sin(h)}{\cos(x)})}{h}$$

$$\displaystyle= \lim_{h \to 0} \frac{-\ln{(\cos h-\tan x \sin h)}}{h}$$

$$\displaystyle=\lim_{h \to 0} \frac{-\ln[({1- \tan x \tanh)(\cos h)}]}{h}$$

$$\displaystyle=\lim_{h \to 0} \frac{-\ln({1- \tan x \tanh)}-\ln{\cos h}}{h}$$

$$\displaystyle=\lim_{h \to 0} -\frac{\ln({1- \tan x \tanh)}}{h}-\lim_{h \to 0} \frac{\ln{\cos h}}{h}$$

$$\displaystyle=\lim_{h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}}{\frac{\tan x \tan h}{h}}-\lim_{h \to 0} {\frac1{\cos h}}{(-\sin h)}$$ (Using L'hopital Rule, since $${\ln{\cos h}}\to 0, {h}\to 0$$)

$$\displaystyle=\lim_{\tan x \tan h \to 0} {\frac{\ln({1+(-\tan x \tan h))}}{-\tan x \tan h}} \lim_{h \to 0} {\frac{\tan x \tan h}{h}}+\lim_{h \to 0} {\tan h} \; (\because h \to 0 \implies \tan h \to 0 \implies \tan x \tan h \to 0)$$

$$\displaystyle=1. \tan x \lim_{h\to 0} \frac{\tan h}{h} +0 \;(\because \lim_{y\to 0} {\ln{(1+y)} \over y}=1)$$

$$\displaystyle= \tan x \lim_{h\to 0} \sec^2{h}=\tan x . 1$$ (Using L'hopital Rule, since $${\tan h}\to 0, {h}\to 0$$)

$$=\tan x$$

I'm not sure if this is in the spirit of the question, but we can also try proving the chain rule in the special case $$(\ln \circ \sec)'(a)=\ln'(\sec a) \cdot \sec'(a)$$. Note that $$\lim_{x \to a}\frac{\ln(\sec x)-\ln(\sec a)}{x-a}=\lim_{x \to a}\frac{\ln(\sec x)-\ln(\sec a)}{\sec x-\sec a} \cdot \lim_{x \to a}\frac{\sec x-\sec a}{x-a} \label{*}\tag{*} \, .$$ For the first limit on the RHS of $$\eqref{*}$$, we can make the substitution $$u=\sec x$$. As $$x\to a$$, $$u\to\sec a$$, and so $$\lim_{u \to \sec a}\frac{\ln(u)-\ln(\sec a)}{u-\sec a}=\ln'(\sec a)=\frac{1}{\sec a}$$ Since the second limit on the RHS is $$\sec'(a)=\sec(a)\tan(a)$$, we get that $$(\ln \circ \sec)'(a)=\tan(a) \, .$$