Question

Derivative of secx with respect to x using first principal.

Answer 1

Hint: Start by converting secx in terms of cosine function followed by the use of the first principle for finding the derivative of the function. Also, you might need to use the formula of cosA-cosB.

Complete step-by-step answer:
If f(x) is any derivable function, then the derivative of f(x), by first principle, is given as:
$\dfrac{d\left( f(x) \right)}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}$
Applying the first principle for finding the derivative of $\sec x$ :
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sec \left( x+h \right)-\sec x}{h}$
Using the formula: $\operatorname{secx}=\dfrac{1}{\cos x}$ , we get
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{\cos \left( x+h \right)}-\dfrac{1}{\cos x}}{h}$
Taking the LCM of the right hand side of the equation as cos(x+h)cosx, we get
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\cos x-\cos \left( x+h \right)}{\cos \left( x+h \right)\cos x}}{h}$
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\cos x-\cos \left( x+h \right)}{h\cos \left( x+h \right)\cos x}$
Using formula: $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$ .
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-2\sin \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{-h}{2} \right)}{h\cos \left( x+h \right)\cos x}$
$\Rightarrow \dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2\sin \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{h}{2} \right)}{h\cos \left( x+h \right)\cos x}$
The above equation can also be written as:
 $\Rightarrow \dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{2x+h}{2} \right)\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}\cos \left( x+h \right)\cos x}$
And in our equation, if we put h=0, $\dfrac{h}{2}=0$ .
We know;
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$ .
Using this to our equation, we get:
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{2x+h}{2} \right)}{\cos \left( x+h \right)\cos x}\times \dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}}$
$\Rightarrow \dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{2x+h}{2} \right)}{\cos \left( x+h \right)\cos x}\times 1$
$\Rightarrow \dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \dfrac{2x+h}{2} \right)}{\cos \left( x+h \right)\cos x}$
Finally putting the limit to the expression, we get:
$\dfrac{d\sec x}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( x \right)}{{{\cos }^{2}}x}$
Using formula: $\tan x=\dfrac{\sin x}{\cos x}\text{ and }\sec x=\dfrac{1}{\cos x}$ .
$\dfrac{d\sec x}{dx}=\sec x\tan x$
Hence, we can say that the derivative of $\sec x$ w.r.t $x$ is \[secx\tan x\] .

Note: Whenever you try to find the derivative avoid using the first principle until and unless it is mentioned in the question. Using the first principle might take longer as compared to the traditional method and may require a good knowledge related to different forms of formulas related to algebra, trigonometry, and also a bit of manipulation.
While solving limits, always keep a habit of checking the indeterminate form of the expression at the beginning of the question. There are a total of eight indeterminate forms that you should be knowing. Also, try to eliminate the particular term, which is causing the expression to be indeterminate.