Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Find the derivatives of sinx

Sagot :

ⲊⲧɑⲅⲊⲏɑᴅⲟᏇ

Answer:

  • Cos x

Solution:

Let f ( x ) = sin x , We need to find f' ( x )

  • We know that

[tex] \quad\Rrightarrow\quad \sf {f'(x) = \lim_{h\to 0 }\dfrac{f(x+h)-r(x)}{h} }[/tex]

Here,

  • f ( x ) = sin x

So, f( x + h ) = sin ( x + h )

  • Putting values

[tex] \implies \tt { f'(x)=\lim_{h\to 0} \dfrac{sin(x+h)-sin\:x}{h}}[/tex]

[tex] \implies \tt {f'(x)= lim_{h\to 0 }\dfrac{2cos\left(\dfrac{x+h+x}{2}\right).sin\left(\dfrac{x+h-x}{2}\right)}{h}\qquad\quad\bigg[ sin\:A - sin\: B = 2cos\left(\dfrac{A+B}{2}\right).sin\left(\dfrac{A-B}{2}\right)\bigg]}[/tex]

[tex] \implies \tt {f'(x)=\lim_{h\to 0 }\dfrac{cos\left(\dfrac{2x+h}{2}\right).sin\left(\dfrac{h}{2}\right) }{h}}[/tex]

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

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

[tex] \implies\tt {f'(x) = \lim_{h\to 0}cos\dfrac{(2x+h)}{2}\times 1 }[/tex]

[tex] \implies\tt {f'(x) = \lim_{h\to 0 }cos \left( \dfrac{2x+h}{2}\right)}[/tex]

[tex] \implies\tt {f'(x) =cos\left( \dfrac{2x+0}{2}\right) }[/tex]

[tex] \implies\tt { f'(x) =cos\left(\dfrac{2x}{2}\right)}[/tex]

[tex] \implies\tt {f'(x) =cos \:x }[/tex]

[tex] \implies\underline{\underline{\pmb{\tt {f'(x) = cos\: x }}} }[/tex]

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.