Find the derivatives of Log(secx) by using first principle

To find the derivative of the function $y=\log (\sec x)$ using the first principle of derivatives, we'll rely on the definition of the derivative as a limit. The first principle states that the derivative of a function $f(x)$ at a point $x$ is given by:

Defining the Derivative

The derivative $f^{\prime }(x)$ can be expressed mathematically as:

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

Applying the Definition

Let's apply this to our function $y=\log (\sec x)$. First, we need to find $f(x+h)$:

So, we have:

\( f(x + h) = \log(\sec(x + h))

Now, we can substitute $f(x)$ and $f(x+h)$ into the limit definition:

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{\log (\sec (x+h))-\log (\sec x)}{h}$

Using Logarithmic Properties

We can utilize the properties of logarithms, specifically that $\log (a)-\log (b)=\log \left(\frac{a}{b}\right)$, to simplify the expression:

\( f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\frac{\sec(x + h)}{\sec x}\right)

Limit Transformation

Next, we can rewrite this limit. We know that $\sec (x)=\frac{1}{\cos (x)}$, so:

\( \frac{\sec(x + h)}{\sec x} = \frac{\cos x}{\cos(x + h)}

This gives us:

\( f'(x) = \lim_{h \to 0} \frac{1}{h} \log\left(\frac{\cos x}{\cos(x + h)}\right)

Using Taylor Series Expansion

To find the limit, we can use the Taylor series expansion for $\cos (x+h)$ around $h=0$:

• $\cos (x+h)\approx \cos x-h\sin x+O(h^2)$

Substituting this approximation into our expression, we get:

$\frac{\cos x}{\cos (x+h)}\approx \frac{\cos x}{\cos x-h\sin x}=\frac{1}{1-\frac{h\sin x}{\cos x}}$

Utilizing the limit property $\log (1+u)\approx u$ for small $u$, we can write:

\( \log\left(\frac{\cos x}{\cos(x + h)}\right) \approx \frac{h \sin x}{\cos x}

Final Steps to Derivative

Now we substitute back into our derivative limit:

f'(x) = \lim_{h \to 0} \frac{1}{h} \cdot \frac{h \sin x}{\cos x} = \frac{\sin x}{\cos x} = \tan x

Conclusion

Thus, we have derived that:

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

This process not only showcases the power of the first principles in calculus but also reinforces our understanding of how derivatives represent the rate of change of functions in a foundational way.