حساب التفاضل والتكامل الأمثلة

y = {(sin (x))}^{cos (x)}

$y = {(\sin (x))}^{\cos (x)}$

خطوة 1

افترض أن

y = f (x)

$y = f (x)$ ، خُذ اللوغاريتم الطبيعي لكلا الطرفين

ln (y) = ln (f (x))

$\ln (y) = \ln (f (x))$ .

ln (y) = ln ({(sin (x))}^{cos (x)})

$\ln (y) = \ln ({(\sin (x))}^{\cos (x)})$

خطوة 2

وسّع

ln ({(sin (x))}^{cos (x)})

$\ln ({(\sin (x))}^{\cos (x)})$ بنقل

cos (x)

$\cos (x)$ خارج اللوغاريتم.

ln (y) = cos (x) ln (sin (x))

$\ln (y) = \cos (x) \ln (\sin (x))$

خطوة 3

أوجِد مشتقة العبارة باستخدام قاعدة السلسلة، مع الأخذ في الاعتبار أن

y

$y$ هو دالة

x

$x$ .

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خطوة 3.1

أوجِد مشتقة الطرف الأيسر

ln (y)

$\ln (y)$ باستخدام قاعدة السلسلة.

\frac{y'}{y} = cos (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \ln (\sin (x))$

خطوة 3.2

أوجِد مشتقة الطرف الأيمن.

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خطوة 3.2.1

أوجِد مشتقة

cos (x) ln (sin (x))

$\cos (x) \ln (\sin (x))$ .

\frac{y'}{y} = \frac{d}{d x} [cos (x) ln (sin (x))]

$\frac{y'}{y} = \frac{d}{d x} [\cos (x) \ln (\sin (x))]$

خطوة 3.2.2

أوجِد المشتقة باستخدام قاعدة الضرب التي تنص على أن

\frac{d}{d x} [f (x) g (x)]

$\frac{d}{d x} [f (x) g (x)]$ هو

f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]

$f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ حيث

f (x) = cos (x)

$f (x) = \cos (x)$ و

g (x) = ln (sin (x))

$g (x) = \ln (\sin (x))$ .

\frac{y'}{y} = cos (x) \frac{d}{d x} [ln (sin (x))] + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) \frac{d}{d x} [\ln (\sin (x))] + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.3

أوجِد المشتقة باستخدام قاعدة السلسلة، والتي تنص على أن

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ هو

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ حيث

f (x) = ln (x)

$f (x) = \ln (x)$ و

g (x) = sin (x)

$g (x) = \sin (x)$ .

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خطوة 3.2.3.1

لتطبيق قاعدة السلسلة، عيّن قيمة

u

$u$ لتصبح

sin (x)

$\sin (x)$ .

\frac{y'}{y} = cos (x) (\frac{d}{d u} [ln (u)] \frac{d}{d x} [sin (x)]) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\sin (x)]) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.3.2

مشتق

ln (u)

$\ln (u)$ بالنسبة إلى

u

$u$ يساوي

\frac{1}{u}

$\frac{1}{u}$ .

\frac{y'}{y} = cos (x) (\frac{1}{u} \frac{d}{d x} [sin (x)]) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) (\frac{1}{u} \frac{d}{d x} [\sin (x)]) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.3.3

استبدِل كافة حالات حدوث

u

$u$ بـ

sin (x)

$\sin (x)$ .

\frac{y'}{y} = cos (x) (\frac{1}{sin (x)} \frac{d}{d x} [sin (x)]) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) (\frac{1}{\sin (x)} \frac{d}{d x} [\sin (x)]) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

\frac{y'}{y} = cos (x) (\frac{1}{sin (x)} \frac{d}{d x} [sin (x)]) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) (\frac{1}{\sin (x)} \frac{d}{d x} [\sin (x)]) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.4

حوّل من

\frac{1}{sin (x)}

$\frac{1}{\sin (x)}$ إلى

csc (x)

$\csc (x)$ .

\frac{y'}{y} = cos (x) (csc (x) \frac{d}{d x} [sin (x)]) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) (\csc (x) \frac{d}{d x} [\sin (x)]) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.5

مشتق

sin (x)

$\sin (x)$ بالنسبة إلى

x

$x$ يساوي

cos (x)

$\cos (x)$ .

\frac{y'}{y} = cos (x) csc (x) cos (x) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos (x) \csc (x) \cos (x) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.6

ارفع

cos (x)

$\cos (x)$ إلى القوة

1

$1$ .

\frac{y'}{y} = {cos}^{1} (x) cos (x) csc (x) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos^{1} (x) \cos (x) \csc (x) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.7

ارفع

cos (x)

$\cos (x)$ إلى القوة

1

$1$ .

\frac{y'}{y} = {cos}^{1} (x) {cos}^{1} (x) csc (x) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos^{1} (x) \cos^{1} (x) \csc (x) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.8

استخدِم قاعدة القوة

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ لتجميع الأُسس.

\frac{y'}{y} = {cos (x)}^{1 + 1} csc (x) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = {\cos (x)}^{1 + 1} \csc (x) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.9

أضف

1

$1$ و

1

$1$ .

\frac{y'}{y} = {cos}^{2} (x) csc (x) + ln (sin (x)) \frac{d}{d x} [cos (x)]

$\frac{y'}{y} = \cos^{2} (x) \csc (x) + \ln (\sin (x)) \frac{d}{d x} [\cos (x)]$

خطوة 3.2.10

مشتق

cos (x)

$\cos (x)$ بالنسبة إلى

x

$x$ يساوي

- sin (x)

$- \sin (x)$ .

\frac{y'}{y} = {cos}^{2} (x) csc (x) + ln (sin (x)) (- sin (x))

$\frac{y'}{y} = \cos^{2} (x) \csc (x) + \ln (\sin (x)) (- \sin (x))$

خطوة 3.2.11

بسّط.

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خطوة 3.2.11.1

أعِد ترتيب الحدود.

\frac{y'}{y} = {cos}^{2} (x) csc (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos^{2} (x) \csc (x) - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.2

بسّط كل حد.

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خطوة 3.2.11.2.1

أعِد كتابة

csc (x)

$\csc (x)$ من حيث الجيوب وجيوب التمام.

\frac{y'}{y} = {cos}^{2} (x) \frac{1}{sin (x)} - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos^{2} (x) \frac{1}{\sin (x)} - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.2.2

اجمع

{cos}^{2} (x)

$\cos^{2} (x)$ و

\frac{1}{sin (x)}

$\frac{1}{\sin (x)}$ .

\frac{y'}{y} = \frac{{cos}^{2} (x)}{sin (x)} - sin (x) ln (sin (x))

$\frac{y'}{y} = \frac{\cos^{2} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))$

\frac{y'}{y} = \frac{{cos}^{2} (x)}{sin (x)} - sin (x) ln (sin (x))

$\frac{y'}{y} = \frac{\cos^{2} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.3

بسّط كل حد.

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خطوة 3.2.11.3.1

أخرِج العامل

cos (x)

$\cos (x)$ من

{cos}^{2} (x)

$\cos^{2} (x)$ .

\frac{y'}{y} = \frac{cos (x) cos (x)}{sin (x)} - sin (x) ln (sin (x))

$\frac{y'}{y} = \frac{\cos (x) \cos (x)}{\sin (x)} - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.3.2

افصِل الكسور.

\frac{y'}{y} = \frac{cos (x)}{1} \cdot \frac{cos (x)}{sin (x)} - sin (x) ln (sin (x))

$\frac{y'}{y} = \frac{\cos (x)}{1} \cdot \frac{\cos (x)}{\sin (x)} - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.3.3

حوّل من

\frac{cos (x)}{sin (x)}

$\frac{\cos (x)}{\sin (x)}$ إلى

cot (x)

$\cot (x)$ .

\frac{y'}{y} = \frac{cos (x)}{1} cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \frac{\cos (x)}{1} \cot (x) - \sin (x) \ln (\sin (x))$

خطوة 3.2.11.3.4

اقسِم

cos (x)

$\cos (x)$ على

1

$1$ .

\frac{y'}{y} = cos (x) cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \cot (x) - \sin (x) \ln (\sin (x))$

\frac{y'}{y} = cos (x) cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \cot (x) - \sin (x) \ln (\sin (x))$

\frac{y'}{y} = cos (x) cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \cot (x) - \sin (x) \ln (\sin (x))$

\frac{y'}{y} = cos (x) cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \cot (x) - \sin (x) \ln (\sin (x))$

\frac{y'}{y} = cos (x) cot (x) - sin (x) ln (sin (x))

$\frac{y'}{y} = \cos (x) \cot (x) - \sin (x) \ln (\sin (x))$

خطوة 4

اعزِل

y'

$y'$ وعوّض بالدالة الأصلية عن

y

$y$ في الطرف الأيمن.

y' = (cos (x) cot (x) - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\cos (x) \cot (x) - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5

بسّط الطرف الأيمن.

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خطوة 5.1

بسّط كل حد.

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خطوة 5.1.1

أعِد كتابة

cot (x)

$\cot (x)$ من حيث الجيوب وجيوب التمام.

y' = (cos (x) \frac{cos (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\cos (x) \frac{\cos (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.1.2

اضرب

cos (x) \frac{cos (x)}{sin (x)}

$\cos (x) \frac{\cos (x)}{\sin (x)}$ .

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خطوة 5.1.2.1

اجمع

cos (x)

$\cos (x)$ و

\frac{cos (x)}{sin (x)}

$\frac{\cos (x)}{\sin (x)}$ .

y' = (\frac{cos (x) cos (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos (x) \cos (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.1.2.2

ارفع

cos (x)

$\cos (x)$ إلى القوة

1

$1$ .

y' = (\frac{{cos}^{1} (x) cos (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos^{1} (x) \cos (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.1.2.3

ارفع

cos (x)

$\cos (x)$ إلى القوة

1

$1$ .

y' = (\frac{{cos}^{1} (x) {cos}^{1} (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos^{1} (x) \cos^{1} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.1.2.4

استخدِم قاعدة القوة

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ لتجميع الأُسس.

y' = (\frac{{cos (x)}^{1 + 1}}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{{\cos (x)}^{1 + 1}}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.1.2.5

أضف

1

$1$ و

1

$1$ .

y' = (\frac{{cos}^{2} (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos^{2} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

y' = (\frac{{cos}^{2} (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos^{2} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

y' = (\frac{{cos}^{2} (x)}{sin (x)} - sin (x) ln (sin (x))) {(sin (x))}^{cos (x)}

$y' = (\frac{\cos^{2} (x)}{\sin (x)} - \sin (x) \ln (\sin (x))) {(\sin (x))}^{\cos (x)}$

خطوة 5.2

طبّق خاصية التوزيع.

y' = \frac{{cos}^{2} (x)}{sin (x)} {sin (x)}^{cos (x)} - sin (x) ln (sin (x)) {sin (x)}^{cos (x)}

$y' = \frac{\cos^{2} (x)}{\sin (x)} {\sin (x)}^{\cos (x)} - \sin (x) \ln (\sin (x)) {\sin (x)}^{\cos (x)}$

خطوة 5.3

اجمع

\frac{{cos}^{2} (x)}{sin (x)}

$\frac{\cos^{2} (x)}{\sin (x)}$ و

{sin (x)}^{cos (x)}

${\sin (x)}^{\cos (x)}$ .

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - sin (x) ln (sin (x)) {sin (x)}^{cos (x)}

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - \sin (x) \ln (\sin (x)) {\sin (x)}^{\cos (x)}$

خطوة 5.4

اضرب

sin (x)

$\sin (x)$ في

{sin (x)}^{cos (x)}

${\sin (x)}^{\cos (x)}$ بجمع الأُسس.

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خطوة 5.4.1

انقُل

{sin (x)}^{cos (x)}

${\sin (x)}^{\cos (x)}$ .

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - ({sin (x)}^{cos (x)} sin (x)) ln (sin (x))

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - ({\sin (x)}^{\cos (x)} \sin (x)) \ln (\sin (x))$

خطوة 5.4.2

اضرب

{sin (x)}^{cos (x)}

${\sin (x)}^{\cos (x)}$ في

sin (x)

$\sin (x)$ .

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خطوة 5.4.2.1

ارفع

sin (x)

$\sin (x)$ إلى القوة

1

$1$ .

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - ({sin (x)}^{cos (x)} {sin}^{1} (x)) ln (sin (x))

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - ({\sin (x)}^{\cos (x)} \sin^{1} (x)) \ln (\sin (x))$

خطوة 5.4.2.2

استخدِم قاعدة القوة

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ لتجميع الأُسس.

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - {sin (x)}^{cos (x) + 1} ln (sin (x))

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - {sin (x)}^{cos (x) + 1} ln (sin (x))

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

y' = \frac{{cos}^{2} (x) {sin (x)}^{cos (x)}}{sin (x)} - {sin (x)}^{cos (x) + 1} ln (sin (x))

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x)}}{\sin (x)} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.5

احذِف العامل المشترك لـ

{sin (x)}^{cos (x)}

${\sin (x)}^{\cos (x)}$ و

sin (x)

$\sin (x)$ .

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خطوة 5.5.1

أخرِج العامل

sin (x)

$\sin (x)$ من

{cos}^{2} (x) {sin (x)}^{cos (x)}

$\cos^{2} (x) {\sin (x)}^{\cos (x)}$ .

y' = \frac{sin (x) ({cos}^{2} (x) {sin (x)}^{cos (x) - 1})}{sin (x)} - {sin (x)}^{cos (x) + 1} ln (sin (x))

$y' = \frac{\sin (x) (\cos^{2} (x) {\sin (x)}^{\cos (x) - 1})}{\sin (x)} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.5.2

ألغِ العوامل المشتركة.

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خطوة 5.5.2.1

اضرب في

1

$1$ .

y' = \frac{sin (x) ({cos}^{2} (x) {sin (x)}^{cos (x) - 1})}{sin (x) \cdot 1} - {sin (x)}^{cos (x) + 1} ln (sin (x))

$y' = \frac{\sin (x) (\cos^{2} (x) {\sin (x)}^{\cos (x) - 1})}{\sin (x) \cdot 1} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.5.2.2

ألغِ العامل المشترك.

$y' = \frac{\sin (x) (\cos^{2} (x) {\sin (x)}^{\cos (x) - 1})}{\sin (x) \cdot 1} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.5.2.3

أعِد كتابة العبارة.

$y' = \frac{\cos^{2} (x) {\sin (x)}^{\cos (x) - 1}}{1} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.5.2.4

اقسِم

$\cos^{2} (x) {\sin (x)}^{\cos (x) - 1}$ على

$1$ .

$y' = \cos^{2} (x) {\sin (x)}^{\cos (x) - 1} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

خطوة 5.6

أعِد ترتيب العوامل في

$\cos^{2} (x) {\sin (x)}^{\cos (x) - 1} - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$ .

$y' = {\sin (x)}^{\cos (x) - 1} \cos^{2} (x) - {\sin (x)}^{\cos (x) + 1} \ln (\sin (x))$

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