Using properties of logarithms

[kinerry]

Here's the problem:

Use logarithmic differentiation to show that if y = tanx, then dy/dx = sec^2x (or (secx)^2 on a calculator)

I know the first step is to rewrite tanx as sinx/cosx, but my trig. isn't so great so I'm at a loss as to the next step.

Any help is appreciated

 

[galactus]

Here's a hint:

\(\displaystyle \L\\\frac{d}{dx}[tan(x)]=\frac{d}{dx}[\frac{sin(x)}{cos(x)}]\\=\frac{cos(x)\frac{d}{dx}[sin(x)]-sin(x)\frac{d}{dx}[cos(x)]}{cos^{2}(x)}\\=\frac{cos(x)cos(x)-sin(x)(-sin(x))}{cos^{2}(x)}\\=\frac{cos^{2}(x)+sin^{2}(x)}{cos^{2}(x)}\\=\frac{1}{cos^{2}}\\=sec^{2}(x)\)

 

[soroban]

Hello, kinerry!

This is a long and silly problem ... but it's good practice.

Use logarithmic differentiation to show that: if \(\displaystyle \,y \,= \,\tan x\), then \(\displaystyle \frac{dy}{dx}\,=\,\sec^2x\)

We have: \(\displaystyle \,y\:=\:\frac{\sin x}{\cos x}\)

Take logs: \(\displaystyle \,\ln(y) \:= \:\ln\left(\frac{\sin x}{\cos x}\right) \:=\:\ln(\sin x)\,-\,\ln(\cos x)\)

Differentiate implicitly: \(\displaystyle \L\,\frac{1}{y}\cdot\frac{dy}{dx} \:=\:\frac{\cos x}{\sin x} - \frac{-\sin x}{\cos }\:=\:\frac{\cos^2x\,+\,\sin^2x}{\sin x\cos x}\:=\:\frac{1}{\sin x\cos x}\)

Then: \(\displaystyle \L\,\frac{dy}{dx}\:=\:y\cdot\frac{1}{\sin x\cos x} \:= \:\tan x\cdot\frac{1}{\sin x\cos x}\;=\;\frac{\sin x}{\cos x}\cdot\frac{1}{\sin x\cos x} \:=\:\frac{1}{\cos^2x}\)

Therefore: \(\displaystyle \L\,\frac{dy}{dx}\:=\:\sec^2x\)

 

[kinerry]

Wow, that was fast
Thank you