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Câu hỏi: Tìm đạo hàm của hàm số sau:
\(y = {{\sin x - x\cos x} \over {\cos x + x\sin x}}.\)
\(y = {{\sin x - x\cos x} \over {\cos x + x\sin x}}.\)
Lời giải chi tiết
$y^{\prime}=\frac{(\sin x-x \cos x)^{\prime}(\cos x+x \sin x)-(\sin x-x \cos x)(\cos x+x \sin x)^{\prime}}{(\cos x+x \sin x)^{2}}$
$=\frac{\left[\cos x-(x \cos x)^{\prime}\right](\cos x+x \sin x)-(\sin x-x \cos x)\left[-\sin x+(x \sin x)^{\prime}\right]}{(\cos x+x \sin x)^{2}}$
$=\frac{[\cos x-(\cos x+x \cdot(-\sin x))](\cos x+x \sin x)-(\sin x-x \cos x)[-\sin x+(\sin x+x \cos x)]}{(\cos x+x \sin x)^{2}}$
$=\frac{(\cos x-\cos x+x \sin x)(\cos x+x \sin x)-(\sin x-x \cos x)(-\sin x+\sin x+x \cos x)}{(\cos x+x \sin x)^{2}}$
$=\frac{x \sin x(\cos x+x \sin x)-(\sin x-x \cos x) \cdot x \cos x}{(\cos x+x \sin x)^{2}}$
$=\frac{x \sin x \cos x+x^{2} \sin ^{2} x-x \sin x \cos x+x^{2} \cos ^{2} x}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2} \sin ^{2} x+x^{2} \cos ^{2} x}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2}\left(\sin ^{2} x+\cos ^{2} x\right)}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2}}{(\cos x+x \sin x)^{2}}$
$y^{\prime}=\frac{(\sin x-x \cos x)^{\prime}(\cos x+x \sin x)-(\sin x-x \cos x)(\cos x+x \sin x)^{\prime}}{(\cos x+x \sin x)^{2}}$
$=\frac{\left[\cos x-(x \cos x)^{\prime}\right](\cos x+x \sin x)-(\sin x-x \cos x)\left[-\sin x+(x \sin x)^{\prime}\right]}{(\cos x+x \sin x)^{2}}$
$=\frac{[\cos x-(\cos x+x \cdot(-\sin x))](\cos x+x \sin x)-(\sin x-x \cos x)[-\sin x+(\sin x+x \cos x)]}{(\cos x+x \sin x)^{2}}$
$=\frac{(\cos x-\cos x+x \sin x)(\cos x+x \sin x)-(\sin x-x \cos x)(-\sin x+\sin x+x \cos x)}{(\cos x+x \sin x)^{2}}$
$=\frac{x \sin x(\cos x+x \sin x)-(\sin x-x \cos x) \cdot x \cos x}{(\cos x+x \sin x)^{2}}$
$=\frac{x \sin x \cos x+x^{2} \sin ^{2} x-x \sin x \cos x+x^{2} \cos ^{2} x}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2} \sin ^{2} x+x^{2} \cos ^{2} x}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2}\left(\sin ^{2} x+\cos ^{2} x\right)}{(\cos x+x \sin x)^{2}}$
$=\frac{x^{2}}{(\cos x+x \sin x)^{2}}$