MTK~SMA/MA
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[tex]\Large\underline{\sf{MATEMATIKA}}[/tex]
Turunan pertama dari
[tex]\mathbf{y=\frac{\sin x}{\sin x+\cos x}}[/tex]
adalah y' = ...
[tex]\mathbf{A.\ \ \frac{\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{B.\ \ \frac{1}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{C.\ \ \frac{2}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{D.\ \ \frac{\sin x-\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{E.\ \ \frac{2\sin x.\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\small\boxed{\tt{Note_{1}=Jawablah\ dengan\ usaha\ sendiri}}[/tex]
[tex]\small\boxed{\tt{Note_{2}=Jaga \ kesehatan \ yah}}[/tex]
Terimakasih ^^
_______________
[tex]\Large\underline{\sf{MATEMATIKA}}[/tex]
Turunan pertama dari
[tex]\mathbf{y=\frac{\sin x}{\sin x+\cos x}}[/tex]
adalah y' = ...
[tex]\mathbf{A.\ \ \frac{\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{B.\ \ \frac{1}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{C.\ \ \frac{2}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{D.\ \ \frac{\sin x-\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\mathbf{E.\ \ \frac{2\sin x.\cos x}{\left(\sin x+\cos x\right)^{2}}}[/tex]
[tex]\small\boxed{\tt{Note_{1}=Jawablah\ dengan\ usaha\ sendiri}}[/tex]
[tex]\small\boxed{\tt{Note_{2}=Jaga \ kesehatan \ yah}}[/tex]
Terimakasih ^^
Penjelasan dengan langkah-langkah:
[tex] \sf y = \frac{sin \: x}{sin \: x + cos \: x'} [/tex]
[tex] \sf v = sin \: x + \: cos \: x \to v'[/tex]
[tex] \sf = cos \: x \: - \: sin \: x[/tex]
[tex] \sf y' = \frac{u'v - uv'}{ {v}^{2} } [/tex]
[tex] \sf = \frac{(cos \: x)(sin \: x + cos \: x) - (sin \: x)(cos \: x - sin \: x)}{ {(sin \: x + cos \: x)}^{2} } [/tex]
[tex] \sf = \frac{sin \: x \: cos \: x \: + {cos \: x }^{2} - sin \: x \: cos \: x \: {sin}^{2} x }{(sin \: x \: + \: cos \: x) {}^{2} } [/tex]
[tex] \bf = \frac{1}{ {(sin \: x \: + \: cos \: x)}^{2} } [/tex]
Opsi ≈ [tex]\tiny\bf{B.\ \ \frac{1}{\left(\sin x+\cos x\right)^{2}}}[/tex]
