Question

Find the derivative of $\dfrac{1}{{\sec x}}$ ?

Answer 1

We will convert the given term in terms of cos. So that we will get rid of fractional terms. Then using the first principle, we will find the derivative of the converted term. If these types of questions are asked to only tick the correct answer, one can answer it if he or she remembers the derivative formula.

Complete step-by-step solution:
We have to find the derivative of $\dfrac{1}{{\sec x}}$
We know that $\cos x = \dfrac{1}{{\sec x}}$
We can convert $\dfrac{1}{{\sec x}}$ in terms of cos
$\Rightarrow \dfrac{1}{{\sec x}} = \cos x$
Now, we have to find the derivative of $\cos x$ .
We know that the first principle is
$\mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$
We will substitute $f\left( x \right) = \cos x$ in the equation, we get
$\mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{\cos \left( {x + h} \right) - \cos \left( x \right)}}{h}$
We also know that $cosa - cosb = 2\sin\left( {\dfrac{{a + b}}{2}} \right)\sin\left( {\dfrac{{b - a}}{2}} \right)$
we will substitute $a = x + h$ and b=x in the equation
$\Rightarrow \cos (x + h) - \cos x = 2\sin\left( {\dfrac{{x + h + x}}{2}} \right)\sin\left( {\dfrac{{x - x - h}}{2}} \right)$
we have used the above formula $\mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \dfrac{{\sin x}}{x} = \mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \dfrac{{\sin \left( {\dfrac{x}{2}} \right)}}{{\dfrac{x}{2}}} = 1$
$= \mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{\cos \left( {x + h} \right) - \cos \left( x \right)}}{h}$
$= \mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{2\sin\left( {\dfrac{{x + h + x}}{2}} \right)\sin\left( {\dfrac{{x - x - h}}{2}} \right)}}{h}$
Simplify the equation
$= \mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{2\sin\left( {x + \dfrac{h}{2}} \right)\sin\left( {\dfrac{{ - h}}{2}} \right)}}{h}$
We divide the numerator and denominator by 2.
$= \mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{\sin\left( {x + \dfrac{h}{2}} \right)\sin\left( {\dfrac{{ - h}}{2}} \right)}}{{\dfrac{h}{2}}}$
We know that $\sin \left(-x\right) = - \sin x$ . Here we substitute $x = \dfrac{h}{2}$
$= - \mathop {\lim }\limits_{h \to 0} {\mkern 1mu} \dfrac{{\sin\left( {x + \dfrac{h}{2}} \right)\sin\left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}$
We know that
$= \mathop { - \lim }\limits_{h \to 0} \sin\left( {x + \dfrac{h}{2}} \right) \times 1$
We know that $\mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \dfrac{{\sin x}}{x} = \mathop {\lim }\limits_{x \to 0} {\mkern 1mu} \dfrac{{\sin \left( {\dfrac{x}{2}} \right)}}{{\dfrac{x}{2}}} = 1$
$= \mathop { - \lim }\limits_{h \to 0} \sin\left( {x + 0} \right) \times 1$
$= - \sin x$
The derivative of $\dfrac{1}{{\sec x}}$ is $- \sin x$.

Note: The derivative of a function is the instantaneous rate of change of the function with respect to the dependent variable. The slope of a function's curve is represented by the first derivative of that function. Limits must be used with caution when determining the value of derivatives. Applying the limits directly will result in an inaccurate solution.