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[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Question ~}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Prove that ~

[tex] \dfrac{d}{dx}\sec(x) = \sec(x) \tan(x) [/tex]


by using first principle of differentiation ~​

Sagot :

Answer:

METHOD I:

(by using the first principle of differentiation)

We have the "Limit definition of Derivatives":

[tex]\boxed{\mathsf{f'(x)= \lim_{h \to 0} \{\frac{f(x+h)-f(x)}{h} \} ....(i)}}[/tex]

Here, f(x) = sec x, f(x+h) = sec (x+h)

  • Substituting these in eqn. (i)

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }[/tex]

  • sec x can be written as 1/ cos(x)

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }[/tex]

  • Taking LCM

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)} \} }[/tex]

  • By Cosines sum to product formula, i.e.,

[tex]\boxed{\mathsf{cos\:A-cos\:B=-2sin(\frac{A+B}{2} )sin(\frac{A-B}{2} )}}[/tex]

=> cos(x) - cos(x+h) = -2sin{(x+x+h)/2}sin{(x-x-h)/2}

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{2sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{sin(\frac{h}{2} )}{h} }[/tex]

  • I shifted a 2 from the first limit to the second limit, since the limits ar ein multiplication this transmission doesn't affect the result

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{2sin(\frac{h}{2} )}{h} }[/tex]

  • 2/ h can also be written as 1/(h/ 2)

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{1\times sin(\frac{h}{2} )}{\frac{h}{2} } }[/tex]

  • We have limₓ→₀ (sin x) / x = 1.

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: 1 }[/tex]

  • h→0 means h/ 2→0

Substituting 0 for h and h/ 2

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)cos(x)} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)}\times \frac{1}{cos x} }[/tex]

  • sin x/ cos x is tan x whereas 1/ cos (x) is sec (x)

[tex]\implies \mathsf{f'(x)= tan(x)\times sec(x) }[/tex]

Hence, we got

[tex]\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}[/tex]

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METHOD II:

(by using other standard derivatives)

[tex] \boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) = \sec x \tan x }}[/tex]

  • sec x can also  be written as (cos x)⁻¹

We have a standard derivative for variables in x raised to an exponent:

[tex] \boxed{ \mathsf{ \frac{d}{dx}(x)^{n} = n(x)^{n - 1} }}[/tex]

Therefore,

[tex] \mathsf{ \frac{d}{dx}( \cos x)^{ - 1} = - 1( \cos \: x) ^{( - 1 - 1} } \\ \implies \mathsf{\ - 1( \cos \: x) ^{- 2 }}[/tex]

  • Any base with negative exponent is equal to its reciprocal with same positive exponent

[tex] \implies \: \mathsf{ - \frac{1}{ (\cos x) {}^{2} } }[/tex]

The process of differentiating doesn't just end here. It follows chain mechanism, I.e.,

while calculating the derivative of a function that itself contains a function, the derivatives of all the inner functions are multiplied to that of the exterior to get to the final result.

  • The inner function that remains is cos x whose derivative is -sin x.

[tex] \implies \mathsf{ - \frac{1}{ (\cos x )^{2} } \times ( - \sin x) }[/tex]

  • cos²x can also be written as (cos x).(cos x)

[tex] \implies \mathsf{ \frac{ \sin x }{ \cos x } \times ( \frac{1}{cos x} ) }[/tex]

  • sin x/ cos x is tan x, while 1/ cos x is sec x

[tex] \implies \mathsf{ \tan x \times \sec x }[/tex]

= sec x. tan x

Hence, Proved!

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