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AnnZ [28]
3 years ago
8

96%AA%E2%96%AA%E2%96%AA%20%20%7B%5Chuge%5Cmathfrak%7BQuestion%20~%7D%7D%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA" id="TexFormula1" title="▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Question ~}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪" alt="▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Question ~}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪" align="absmiddle" class="latex-formula">
Prove that ~

\dfrac{d}{dx}\sec(x) = \sec(x) \tan(x)


by using first principle of differentiation ~​
Mathematics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

<h3>METHOD I:</h3>

(by using the first principle of differentiation)

We have the <u>"Limit definition of Derivatives"</u>:

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

<em>Here, f(x) = sec x, f(x+h) = sec (x+h)</em>

  • <em>Substituting these in eqn. (i)</em>

\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }<em />

  • <em>sec x can be written as 1/ cos(x)</em>

<em />\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }<em />

  • <em>Taking LCM</em>

<em />\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)}  \} }<em />

  • <em>By Cosines sum to product formula, i.e.,</em>

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

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

\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}   }

  • <em>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</em>

\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}   }

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

\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} }   }

  • <em>We have limₓ→₀ (sin x) / x = 1. </em>

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

  • <em>h→0 means h/ 2→0</em>

<em>Substituting 0 for h and h/ 2</em>

<em />\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }<em />

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

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

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

\implies \mathsf{f'(x)=  tan(x)\times sec(x)  }

Hence, we got

\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}

-  - - - -  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>METHOD II:</h3>

(by using other standard derivatives)

\boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) =  \sec x   \tan x }}

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

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

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

Therefore,

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

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

\implies \: \mathsf{  - \frac{1}{ (\cos x)  {}^{2} } }

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

<em>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</em>.

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

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

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

\implies \mathsf{   \frac{ \sin x }{ \cos x   }  \times (  \frac{1}{cos x} )   }

  • <u>sin x/ cos x</u> is tan x, while <u>1/ cos x</u> is sec x

\implies \mathsf{    \tan x  \times  \sec x  }

= sec x. tan x

<h3>Hence, Proved!</h3>
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