Sin 2x'in Türevi

f(x) = sin 2x ⇒ f'(x) = 2.cos 2x'tir.

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sin2x türevi.png

Sin 2x'in Türevi Nedir ?

Sin 2x'in türevi, 2.cos 2x'tir.

$\frac{d}{d x} (sin 2 x) = 2. cos 2 x$

Sin 2x'in Türevinin İspatı

1. Yol

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 ( x + h ) - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin ( 2 x + 2 h ) - sin 2 x}{h}$

$s i n (p + q) = s i n p . c o s q + s i n q . c o s p$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . cos 2 h + sin 2 h . cos 2 x - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . cos 2 h - sin 2 x + sin 2 h . cos 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x}{h} . \frac{2}{2}]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. [ sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x ]}{2. h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin 2 x . ( cos 2 h - 1 ) + 2. sin 2 h . cos 2 x}{2 h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{2. sin 2 x . ( cos 2 h - 1 )}{2 h} + \frac{2. sin 2 h . cos 2 x}{2 h}]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin 2 x . ( cos 2 h - 1 )}{2 h} + h \to 0 lim \frac{2. sin 2 h . cos 2 x}{2 h}$

$(sin 2 x)^{'} = 2. sin 2 x . h \to 0 lim \frac{cos 2 h - 1}{2 h} + 2. cos 2 x . h \to 0 lim \frac{sin 2 h}{2 h}$

$h = 2 h$ $(h \to 0)$

$(sin 2 x)^{'} = 2. sin 2 x . h \to 0 lim \frac{cos h - 1}{h} + 2. cos 2 x . h \to 0 lim \frac{sin h}{h}$

$t \to 0 l i m \frac{s i n t}{t} = 1$ $t \to 0 l i m \frac{c o s t - 1}{t} = 0$

$(sin 2 x)^{'} = 2. sin 2 x .0 + 2. cos 2 x .1$

$(sin 2 x)^{'} = 0 + 2. cos 2 x$

$(sin 2 x)^{'} = 2. cos 2 x$

2. Yol

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 ( x + h ) - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin ( 2 x + 2 h ) - sin 2 x}{h}$

$s i n p - s i n q = 2. s i n (\frac{p - q}{2}) . c o s (\frac{p + q}{2})$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 x + 2 h - 2 x}{2} ) . cos ( \frac{2 x + 2 h + 2 x}{2} )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 h}{2} ) . cos ( \frac{4 x + 2 h}{2} )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 . h}{2} ) . cos [ \frac{2 . ( 2 x + h )}{2} ]}{h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{2. sin h}{h} . cos (2 x + h)]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin h}{h} . h \to 0 lim cos (2 x + h)$

$(sin 2 x)^{'} = 2. h \to 0 lim \frac{sin h}{h} . h \to 0 lim cos (2 x + h)$

$t \to 0 l i m \frac{s i n t}{t} = 1$

$(sin 2 x)^{'} = 2.1. cos (2 x + 0)$

$(sin 2 x)^{'} = 2.1. cos 2 x$

$(sin 2 x)^{'} = 2. cos 2 x$

3. Yol

Sin 2x ve cos 2x fonksiyonlarının sonsuz seri şeklindeki açılımlarından faydalanarak da sin 2x'in türevinin 2.cos 2x'e eşit olduğunu ispatlayabiliriz. Sin 2x ve cos 2x fonksiyonlarının sonsuz seri şeklindeki açılımları aşağıdaki gibidir.

$s i n 2 x = 2 x - \frac{( 2 x ) ^{3}}{3 !} + \frac{( 2 x ) ^{5}}{5 !} - \frac{( 2 x ) ^{7}}{7 !} + \frac{( 2 x ) ^{9}}{9 !} - ...$

$c o s 2 x = 1 - \frac{( 2 x ) ^{2}}{2 !} + \frac{( 2 x ) ^{4}}{4 !} - \frac{( 2 x ) ^{6}}{6 !} + \frac{( 2 x ) ^{8}}{8 !} - ...$

$sin 2 x = 2 x - \frac{( 2 x ) ^{3}}{3 !} + \frac{( 2 x ) ^{5}}{5 !} - \frac{( 2 x ) ^{7}}{7 !} + \frac{( 2 x ) ^{9}}{9 !} - ...$

$(sin 2 x)^{'} = [2 x - \frac{( 2 x ) ^{3}}{3 !} + \frac{( 2 x ) ^{5}}{5 !} - \frac{( 2 x ) ^{7}}{7 !} + \frac{( 2 x ) ^{9}}{9 !} - ...]^{'}$

$(sin 2 x)^{'} = (2 x)^{'} - [\frac{( 2 x ) ^{3}}{3 !}]^{'} + [\frac{( 2 x ) ^{5}}{5 !}]^{'} - [\frac{( 2 x ) ^{7}}{7 !}]^{'} + [\frac{( 2 x ) ^{9}}{9 !}]^{'} - ...$

$(sin 2 x)^{'} = 2 - \frac{3. ( 2 x ) ^{2} . ( 2 x ) ^{'}}{3 !} + \frac{5. ( 2 x ) ^{4} . ( 2 x ) ^{'}}{5 !} - \frac{7. ( 2 x ) ^{6} . ( 2 x ) ^{'}}{7 !} + \frac{9. ( 2 x ) ^{8} . ( 2 x ) ^{'}}{9 !} - ...$

$(sin 2 x)^{'} = 2 - \frac{3. ( 2 x ) ^{2} .2}{3 !} + \frac{5. ( 2 x ) ^{4} .2}{5 !} - \frac{7. ( 2 x ) ^{6} .2}{7 !} + \frac{9. ( 2 x ) ^{8} .2}{9 !} - ...$

$(sin 2 x)^{'} = 2 - \frac{3 .2. ( 2 x ) ^{2}}{3 .2 !} + \frac{5 .2. ( 2 x ) ^{4}}{5 .4 !} - \frac{7 .2. ( 7 x ) ^{6}}{7 .6 !} + \frac{9 .2. ( 2 x ) ^{8}}{9 .8 !} - ...$

$(sin 2 x)^{'} = 2. [1 - \frac{( 2 x ) ^{2}}{2 !} + \frac{( 2 x ) ^{4}}{4 !} - \frac{( 2 x ) ^{6}}{6 !} + \frac{( 2 x ) ^{8}}{8 !} - ...]$

$(sin 2 x)^{'} = 2. cos 2 x$

# türev # sinüs # trigonometri

Published Date:

January 21, 2021

Updated Date:

July 19, 2024

Sin 2x'in Türeviself.__wrap_n!=1&&self.__wrap_b(":R35nlttada:",1)

Sin 2x'in Türevi Nedir ?

Sin 2x'in Türevinin İspatı

1. Yol

f′(x)=h→0lim​hf(x+h)−f(x)​

(sin2x)′=h→0lim​hsin2(x+h)−sin2x​

(sin2x)′=h→0lim​hsin(2x+2h)−sin2x​

sin(p+q)=sinp.cosq+sinq.cosp​

(sin2x)′=h→0lim​hsin2x.cos2h+sin2h.cos2x−sin2x​

(sin2x)′=h→0lim​hsin2x.cos2h−sin2x+sin2h.cos2x​

(sin2x)′=h→0lim​hsin2x.(cos2h−1)+sin2h.cos2x​

(sin2x)′=h→0lim​[hsin2x.(cos2h−1)+sin2h.cos2x​.22​]

(sin2x)′=h→0lim​2.h2.[sin2x.(cos2h−1)+sin2h.cos2x]​

(sin2x)′=h→0lim​2h2.sin2x.(cos2h−1)+2.sin2h.cos2x​

(sin2x)′=h→0lim​[2h2.sin2x.(cos2h−1)​+2h2.sin2h.cos2x​]

(sin2x)′=h→0lim​2h2.sin2x.(cos2h−1)​+h→0lim​2h2.sin2h.cos2x​

2. Yol

f′(x)=h→0lim​hf(x+h)−f(x)​

(sin2x)′=h→0lim​hsin2(x+h)−sin2x​

(sin2x)′=h→0lim​hsin(2x+2h)−sin2x​

sinp−sinq=2.sin(2p−q​).cos(2p+q​)​

(sin2x)′=h→0lim​h2.sin(22x+2h−2x​).cos(22x+2h+2x​)​

(sin2x)′=h→0lim​h2.sin(22h​).cos(24x+2h​)​

(sin2x)′=h→0lim​h2.sin(2​2​.h​).cos[2​2​.(2x+h)​]​

(sin2x)′=h→0lim​[h2.sinh​.cos(2x+h)]

(sin2x)′=h→0lim​h2.sinh​.h→0lim​cos(2x+h)

(sin2x)′=2.h→0lim​hsinh​.h→0lim​cos(2x+h)

3. Yol

Sin 2x'in Türevi

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 ( x + h ) - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin ( 2 x + 2 h ) - sin 2 x}{h}$

$s i n (p + q) = s i n p . c o s q + s i n q . c o s p$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . cos 2 h + sin 2 h . cos 2 x - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . cos 2 h - sin 2 x + sin 2 h . cos 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x}{h} . \frac{2}{2}]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. [ sin 2 x . ( cos 2 h - 1 ) + sin 2 h . cos 2 x ]}{2. h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin 2 x . ( cos 2 h - 1 ) + 2. sin 2 h . cos 2 x}{2 h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{2. sin 2 x . ( cos 2 h - 1 )}{2 h} + \frac{2. sin 2 h . cos 2 x}{2 h}]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin 2 x . ( cos 2 h - 1 )}{2 h} + h \to 0 lim \frac{2. sin 2 h . cos 2 x}{2 h}$

$f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin 2 ( x + h ) - sin 2 x}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{sin ( 2 x + 2 h ) - sin 2 x}{h}$

$s i n p - s i n q = 2. s i n (\frac{p - q}{2}) . c o s (\frac{p + q}{2})$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 x + 2 h - 2 x}{2} ) . cos ( \frac{2 x + 2 h + 2 x}{2} )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 h}{2} ) . cos ( \frac{4 x + 2 h}{2} )}{h}$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin ( \frac{2 . h}{2} ) . cos [ \frac{2 . ( 2 x + h )}{2} ]}{h}$

$(sin 2 x)^{'} = h \to 0 lim [\frac{2. sin h}{h} . cos (2 x + h)]$

$(sin 2 x)^{'} = h \to 0 lim \frac{2. sin h}{h} . h \to 0 lim cos (2 x + h)$

$(sin 2 x)^{'} = 2. h \to 0 lim \frac{sin h}{h} . h \to 0 lim cos (2 x + h)$