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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]

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