Yogendra Kumar
Last Activity: 3 Years ago
To find the derivative of the function 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 at a point is given by:
Defining the Derivative
The derivative can be expressed mathematically as:
Applying the Definition
Let's apply this to our function . First, we need to find :
So, we have:
\( f(x + h) = \log(\sec(x + h))
Now, we can substitute and into the limit definition:
Using Logarithmic Properties
We can utilize the properties of logarithms, specifically that , 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 , 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 around :
Substituting this approximation into our expression, we get:
Utilizing the limit property for small , 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.