Math Doubts

Derivative of secx formula Proof

$x$ is angle of right angled triangle and secant function is written as $\sec{x}$. The differentiation of $\sec{x}$ function with respect to $x$ is written as $\dfrac{d}{dx} \sec{x}$ in differential calculus.

Express Differentiation of function in Limit form

According to principal relation of differentiation of a unction in limit form, the differentiation of secant function can be expressed in limit from to derive derivative of $\sec{x}$ with respect to $x$ in differential calculus.

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

Take $f(x) = \sec{x}$, then $f(x) = \sec{(x+h)}$. Substitute them in the differentiation rule in limit form.

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

Transform secant functions in terms of cos functions

There is no special trigonometric identities in terms of secant functions. So, each secant function should be transform in other form. Mathematically, it is possible to convert each secant function in terms of cosine function as per reciprocal relation of secant and cosine functions.

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

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

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

Use Sum to Product Trigonometric identity

The cosine of sum to product transformation identity simplifies the numerator by transforming sum of the cosine terms in terms of product of sine functions.

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

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

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

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

Simplify expression by negative trigonometric identity

According to negative trigonometric identity, sine of negative angle is equal to negative of sine of angle.

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

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

Split the expression as multiplying factors

Part of the trigonometric expression represent lim sinx/x as x approaches 0 formula. So, split the trigonometric expression as two multiplying factors.

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

In order to apply lim sinx/x as x approaches 0 formula, the denominator should be same as the angle in the sin function. So, shift number 2 from numerator to denominator.

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

Apply Limit Product Rule

The value of limit belongs to both multiplying factors. So, it can be written to both multiplying factors.

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

If $h \to 0$ then $\dfrac{h}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{h}{2} \to 0$. Change limit of second function but no need to change the limit for first multiplying function.

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

Obtain derivative of secx function

Substitute, $h = 0$ in first multiplying factor and the second multiplying factor represents lim sinx/x as x approaches 0 formula. So, its value is equal to one.

$\implies$ $\dfrac{d}{dx} \sec{x}$ $\,=\,$ $\dfrac{\sin{\Bigg[\dfrac{2x+0}{2}\Bigg]}}{\cos{(x+0)}\cos{x}} \times 1$

$\implies$ $\dfrac{d}{dx} \sec{x}$ $\,=\,$ $\dfrac{\sin{\Bigg[\dfrac{2x}{2}\Bigg]}}{\cos{x}\cos{x}}$

$\implies$ $\dfrac{d}{dx} \sec{x}$ $\,=\,$ $\require{cancel} \dfrac{\sin{\Bigg[\dfrac{\cancel{2}x}{\cancel{2}}\Bigg]}}{\cos{x}\cos{x}}$

$\implies$ $\dfrac{d}{dx} \sec{x}$ $\,=\,$ $\dfrac{\sin{x}}{\cos{x}\cos{x}}$

The trigonometric function can be divided as two multiplying factors.

$\implies$ $\dfrac{d}{dx} \sec{x}$ $\,=\,$ $\dfrac{1}{\cos{x}} \times \dfrac{\sin{x}}{\cos{x}}$

As per reciprocal identity of cosine function, the reciprocal of cos function is written as secant function and the quotient of sine by cosine functions is equal to tan function as per quotient identity of sine and cosine functions.

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

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

Therefore, it is proved that the derivative of $\sec{x}$ with respect to $x$ is equal to the product of $\sec{x}$ and $\tan{x}$ functions.



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