Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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 choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.