Math Doubts

Derivative of secx Proof

The derivative of secant function with respect to a variable is equal to the product of secant and tangent functions. If $x$ is considered to represents a variable, then the secant function is written in mathematical form as $\sec{x}$. The differentiation of the $\sec{x}$ with respect to $x$ is equal to the product of $\sec{x}$ and $\tan{x}$. According to the first principle of differentiation, the derivative of secant function is derived in differential calculus.

Differentiation of function in Limit form

Firstly, write the derivative of a function in terms of $x$ in limits operation form according to the definition of the derivative.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

Now, take $\Delta x = h$ and write the equation in terms of $h$ instead of $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

If $f{(x)} = \sec{x}$, then $f{(x+h)} = \sec{(x+h)}$. Now, let’s start proving the differentiation of $\sec{x}$ function with respect to $x$ by first principle.

$\implies$ $\dfrac{d}{dx}{\, (\sec{x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sec{(x+h)}-\sec{x}}{h}}$

Simplify the entire function

According to reciprocal identity of cosine function, the difference of the secant functions in the numerator can be simplified by converting them into cosine functions .

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{1}{\cos{(x+h)}}-\dfrac{1}{\cos{x}}}{h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{\cos{x}-\cos{(x+h)}}{\cos{(x+h)}\cos{x}}}{h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\cos{x}-\cos{(x+h)}}{h\cos{(x+h)}\cos{x}}}$

The difference of cosine functions can be combined by using difference to product trigonometric identity of cos functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{x+x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-(x+h)}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{x-x-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\require{cancel} \large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{\cancel{x}-\cancel{x}-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{-h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

The sine of negative angle is equal to negative of sine of angle as per even/ odd trigonometric identity of sine function.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{-2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\Bigg(-\sin{\Bigg[\dfrac{h}{2}\Bigg]\Bigg)}}{h\cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h\cos{(x+h)}\cos{x}}}$

It is essential to split the trigonometric function as product of two functions.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Bigg[\dfrac{2x+h}{2}\Bigg]} \times \sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h \times \cos{(x+h)}\cos{x}}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]} \times 2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\cos{(x+h)}\cos{x} \times h}}$

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}} \times \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}\Bigg)}$

By using product rule of limits, the limit of product of two functions can be evaluated by calculating the product of their limits.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{2\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{h}}$

In numerator of second factor, there is a number $2$. It actually multiplies the other factor and it divides the function in denominator. So, shift it to denominator.

$=\,\,\,$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{2x+h}{2}\Bigg]}}{\cos{(x+h)}\cos{x}}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

Evaluate Limits of trigonometric functions

Let us first find the limit of the first trigonometric function by the direct substitution method but do not disturb the second function.

$=\,\,\,$ $\dfrac{\sin{\Bigg[\dfrac{2x+0}{2}\Bigg]}}{\cos{(x+0)}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{\Bigg[\dfrac{2x}{2}\Bigg]}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\require{cancel} \dfrac{\sin{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\dfrac{\sin{x}}{\cos{x}} \times \dfrac{1}{\cos{x}}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\tan{x} \times \sec{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\tan{x}\sec{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{h \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

The second function is similar to the limit of $\dfrac{\sin{x}}{x}$ as $x$ approaches $0$ formula but the input is slightly different. So, try to adjust it to obtain the form same as the limit formula of trigonometric function.

If $h \to 0$, then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. It has proved that if $h$ approaches zero, then $\dfrac{h}{2}$ also approaches zero.

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{\frac{h}{2} \,\to\, 0}{\normalsize \dfrac{\sin{\Bigg[\dfrac{h}{2}\Bigg]}}{\dfrac{h}{2}}}$

Take $z = \dfrac{h}{2}$ and express the limit of trigonometric function in terms of $z$.

$=\,\,\,$ $\sec{x}\tan{x}$ $\times$ $\large \displaystyle \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$

According to limit of sinx/x as x approaches 0 formula, the limit of the trigonometric function is equal to one.

$=\,\,\,$ $\sec{x}\tan{x} \times 1$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \sec{x}} \,=\, \sec{x}\tan{x}$

Therefore, it is proved that the derivative of secant function with respect to a variable is equal to the product of secant and tangent functions.

Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more